Complete the Square to Find Extrema - Algebra 1

The area of a rectangle is given by $A(x) = x^2 - 10x + 30$ , where $x$ is the length of one side in feet. What is the minimum area of this rectangle?
Answer: 5 square feet When you see a quadratic function representing area and need to find a minimum value, you're looking for the vertex of a parabola that opens upward. Since the coefficient of $x^2$ is positive (1), this parabola has a minimum point at its vertex. To find the minimum area, you can complete the square or use the vertex formula. Let's complete the square for $A(x) = x^2 - 10x + 30$ : $A(x) = x^2 - 10x + 30$ $A(x) = (x^2 - 10x + 25) + 30 - 25$ $A(x) = (x - 5)^2 + 5$ This vertex form shows the minimum occurs when $x = 5$ , giving a minimum area of 5 square feet. Looking at the wrong answers: Choice A (25 square feet) likely comes from incorrectly thinking the minimum is at $x = 0$ and calculating $A(0) = 30$ , then somehow getting confused with the algebra. Choice B (10 square feet) might result from using $x = 10$ (the coefficient of the linear term) instead of finding the actual vertex. Choice D (30 square feet) is the y-intercept when $x = 0$ , but this isn't the minimum value. Remember that for any quadratic $ax^2 + bx + c$ where $a > 0$ , the minimum occurs at $x = -\frac{b}{2a}$ . Here, that's $x = -\frac{(-10)}{2(1)} = 5$ . Always convert to vertex form or use the vertex formula when finding maximum or minimum values of quadratic functions.
Complete the square to find the vertex of the parabola $f(x)=x^2-8x+3$ .
Answer: $(4,-13)$ This question tests your understanding of completing the square to rewrite a quadratic in vertex form, which reveals the maximum or minimum value. Completing the square transforms a quadratic from standard form (ax² + bx + c) into vertex form (a(x - h)² + k), which immediately shows us the vertex at (h, k)—the highest or lowest point on the parabola—without any calculation needed once we're in that form. Starting with f(x) = x² - 8x + 3, we complete the square on the first two terms: take half of -8 to get -4, square it to get 16, then add and subtract inside: f(x) = (x² - 8x + 16) - 16 + 3. The first part is a perfect square (x - 4)², and simplifying the constants: -16 + 3 = -13. So vertex form is f(x) = (x - 4)² - 13, with vertex at (4, -13)! Choice C is correct because it properly completes the square to get vertex form f(x) = (x - 4)² - 13, revealing the vertex at (4, -13). Excellent work! Choice A has the y-coordinate wrong as 19 instead of -13, which might happen if you added 16 instead of subtracting it when completing the square. When we add 16 to complete the square, we must also subtract 16 to keep the expression equal! Quick memory trick: in vertex form a(x - h)² + k, the signs are tricky! The vertex x-coordinate is h, but in the formula it appears as 'minus h,' so (x - 3)² has vertex at x = 3, while (x + 3)² has vertex at x = -3. The sign flips! But k is straightforward—it's just the constant at the end.
What is the minimum value of $f(x)=x^2+6x+2$ ? (You may use completing the square.)
Answer: $-7$ This question tests your understanding of completing the square to rewrite a quadratic in vertex form, which reveals the maximum or minimum value. To complete the square for x² + bx, we add and subtract (b/2)²—half the middle coefficient, squared. This creates a perfect square trinomial (x + b/2)² and leaves us with extra terms to simplify, giving us the vertex form that shows the vertex clearly. Starting with f(x) = x² + 6x + 2, we complete the square on the first two terms: take half of 6 to get 3, square it to get 9, then add and subtract inside: f(x) = (x² + 6x + 9) - 9 + 2. The first part is a perfect square (x + 3)², and simplifying the constants: -9 + 2 = -7. So vertex form is f(x) = (x + 3)² - 7, with vertex at (-3, -7)! Choice B is correct because it properly completes the square to get vertex form f(x) = (x + 3)² - 7, revealing the vertex at (-3, -7) with minimum value of -7. Excellent work! Choice A gives 2, which is just the original constant term, not the minimum value. The minimum occurs at the vertex, which we find by completing the square—it's not just the constant term from the original equation! Here's the completing-the-square recipe: (1) Make sure the x² term has coefficient 1 (factor out a if needed), (2) Take half of the x coefficient and square it, (3) Add and subtract that perfect square, (4) Rewrite the first three terms as a perfect square binomial, (5) Simplify the constant terms. Follow these steps and you'll get to vertex form every time!
A farmer wants to fence a rectangular garden. The area function is $A(w) = w(24-2w) = -2w^2 + 24w$ , where $w$ is the width in feet. What width maximizes the area, and what is the maximum area?
Answer: 6 feet; 72 square feet When you see a quadratic function representing a real-world maximum or minimum problem, you're looking for the vertex of a parabola. Since this area function $A(w) = -2w^2 + 24w$ has a negative coefficient for $w^2$ , it opens downward, meaning the vertex gives you the maximum value. To find the width that maximizes area, use the vertex formula. For a quadratic $f(x) = ax^2 + bx + c$ , the x-coordinate of the vertex is $x = -\frac{b}{2a}$ . Here, $a = -2$ and $b = 24$ , so: $w = -\frac{24}{2(-2)} = -\frac{24}{-4} = 6$ The maximum width is 6 feet. To find the maximum area, substitute $w = 6$ into the function: $A(6) = -2(6)^2 + 24(6) = -72 + 144 = 72$ square feet Choice B (6 feet; 72 square feet) is correct. Choice A incorrectly uses 12 feet as the width, which is actually where the area would be zero (when $24 - 2w = 0$ ). Choice C uses 8 feet, which exceeds the optimal width and would give a smaller area of 64 square feet. Choice D has the correct width but calculates the wrong area—likely from an arithmetic error in the substitution. Strategy tip: For quadratic optimization problems, always use the vertex formula $x = -\frac{b}{2a}$ to find the optimal input value, then substitute back to find the optimal output. Double-check your arithmetic in the final substitution step, as this is where many errors occur.
Complete the square to find the vertex of the quadratic function $f(x)=x^2+8x+3$ .
Answer: $(-4,-13)$ This question tests your understanding of completing the square to rewrite a quadratic in vertex form, which reveals the maximum or minimum value. Completing the square transforms a quadratic from standard form (ax² + bx + c) into vertex form (a(x - h)² + k), which immediately shows us the vertex at (h, k)—the highest or lowest point on the parabola—without any calculation needed once we're in that form. Starting with f(x) = x² + 8x + 3, we complete the square on the first two terms: take half of 8 to get 4, square it to get 16, then add and subtract inside: f(x) = (x² + 8x + 16) - 16 + 3. The first part is a perfect square (x + 4)², and simplifying the constants: -16 + 3 = -13. So vertex form is f(x) = (x + 4)² - 13, with vertex at (-4, -13)! Choice B is correct because it properly completes the square to get vertex form (x + 4)² - 13, revealing the vertex at (-4, -13) with minimum value of -13. Excellent work! Choice A has a sign error in the vertex form: when we write (x - h)², a positive h means (x - positive number), but a negative h means (x - negative number) which looks like (x + positive number). This choice has the sign flipped! In vertex form, the vertex x-coordinate has the opposite sign from what appears in the parentheses. Here's the completing-the-square recipe: (1) Make sure the x² term has coefficient 1 (factor out a if needed), (2) Take half of the x coefficient and square it, (3) Add and subtract that perfect square, (4) Rewrite the first three terms as a perfect square binomial, (5) Simplify the constant terms. Follow these steps and you'll get to vertex form every time! Remember why we're doing this: vertex form isn't just busywork—it instantly tells you the most important point on the parabola (the vertex) without graphing or calculating. In real-world problems, that vertex often represents the best or worst outcome, like maximum profit or minimum cost. Powerful stuff!
Complete the square to find the vertex of $g(x)=-2x^2+8x-1$ .
Answer: $(2,7)$ This question tests your understanding of completing the square to rewrite a quadratic in vertex form, which reveals the maximum or minimum value. Completing the square transforms a quadratic from standard form (ax² + bx + c) into vertex form (a(x - h)² + k), which immediately shows us the vertex at (h, k)—the highest or lowest point on the parabola—without any calculation needed once we're in that form. Starting with g(x) = -2x² + 8x - 1, we first factor out -2 from the x terms: g(x) = -2(x² - 4x) - 1. Now complete the square inside: take half of -4 to get -2, square it to get 4, so g(x) = -2(x² - 4x + 4 - 4) - 1 = -2((x - 2)² - 4) - 1 = -2(x - 2)² + 8 - 1 = -2(x - 2)² + 7. Choice A is correct because it properly completes the square to get vertex form g(x) = -2(x - 2)² + 7, revealing the vertex at (2, 7). Excellent work! Choice B gives the vertex as (2, -9), which might happen if you made an error with the signs when distributing the -2. When we have -2 times -4, we get +8, not -8. Sign errors are common but avoidable with careful work! The easiest completing the square problems have even middle coefficients: x² + 6x is easier than x² + 7x because half of 6 is 3 (nice and clean), and 3² = 9. Look for these easier patterns when practicing! Remember why we're doing this: vertex form isn't just busywork—it instantly tells you the most important point on the parabola (the vertex) without graphing or calculating. In real-world problems, that vertex often represents the best or worst outcome, like maximum profit or minimum cost. Powerful stuff!
At what $x$ -value does $f(x)=-(x^2-10x+4)$ reach its maximum? (Use completing the square.)
Answer: $x=5$ This question tests your understanding of completing the square to rewrite a quadratic in vertex form, which reveals the maximum or minimum value. Completing the square transforms a quadratic from standard form (ax² + bx + c) into vertex form (a(x - h)² + k), which immediately shows us the vertex at (h, k)—the highest or lowest point on the parabola—without any calculation needed once we're in that form. Starting with f(x) = -(x² - 10x + 4), we can complete the square inside the parentheses: take half of -10 to get -5, square it to get 25, then f(x) = -(x² - 10x + 25 - 25 + 4) = -((x - 5)² - 21) = -(x - 5)² + 21. This function is in vertex form: y = -(x - 5)² + 21. Reading from this form, the vertex is at (5, 21). Since the coefficient of the squared term is negative, the parabola opens down, making this vertex a maximum. The maximum value is 21, and it occurs when x = 5. Choice C is correct because it properly completes the square to get vertex form f(x) = -(x - 5)² + 21, revealing the vertex at (5, 21) with x = 5 being where the maximum occurs. Excellent work! Choice A gives x = -5, which has a sign error. In vertex form -(x - 5)², the vertex x-coordinate is 5 (positive), not -5. Remember: the sign in the parentheses is opposite to the actual x-coordinate of the vertex! Quick memory trick: in vertex form a(x - h)² + k, the signs are tricky! The vertex x-coordinate is h, but in the formula it appears as 'minus h,' so (x - 3)² has vertex at x = 3, while (x + 3)² has vertex at x = -3. The sign flips! But k is straightforward—it's just the constant at the end.
The height of a projectile is modeled by $h(t) = -4t^2 + 16t + 12$ , where $t$ is time in seconds. At what time does the projectile reach its maximum height?
Answer: 2 seconds When you see a quadratic function modeling projectile motion, you're looking for the vertex of a parabola. Since the coefficient of $t^2$ is negative (-4), this parabola opens downward, making the vertex the maximum point. For any quadratic in the form $h(t) = at^2 + bt + c$ , the time at which the maximum (or minimum) occurs is given by $t = -\frac{b}{2a}$ . In your equation $h(t) = -4t^2 + 16t + 12$ , you have $a = -4$ and $b = 16$ . Substituting into the vertex formula: $t = -\frac{16}{2(-4)} = -\frac{16}{-8} = 2$ seconds. You can verify this by checking that the height increases before $t = 2$ and decreases after. At $t = 1$ : $h(1) = -4 + 16 + 12 = 24$ . At $t = 2$ : $h(2) = -16 + 32 + 12 = 28$ . At $t = 3$ : $h(3) = -36 + 48 + 12 = 24$ . The maximum height of 28 occurs at $t = 2$ . Looking at the wrong answers: (A) 8 seconds comes from incorrectly using $t = -\frac{a}{2b}$ instead of $t = -\frac{b}{2a}$ . (B) 4 seconds results from forgetting the negative sign: $\frac{16}{2(4)} = 2$ , but using positive 4 instead of negative 4. (D) 16 seconds likely comes from confusing the coefficient $b = 16$ with the time value. Remember: for projectile problems, always use the vertex formula $t = -\frac{b}{2a}$ to find when maximum height occurs.
What is the minimum value of $f(x) = x^2 + 6x + 2$ ? (Find it by completing the square.)
Answer: $-7$ This question tests your understanding of completing the square to rewrite a quadratic in vertex form, which reveals the maximum or minimum value. The vertex of a parabola is its extreme point: if the parabola opens upward (a > 0), the vertex is the minimum (lowest point), and if it opens downward (a < 0), the vertex is the maximum (highest point). The y-coordinate of the vertex, k, is that extreme value. Starting with f(x) = x² + 6x + 2, we complete the square on the first two terms: take half of 6 to get 3, square it to get 9, then add and subtract inside: f(x) = (x² + 6x + 9) - 9 + 2. The first part is a perfect square (x + 3)², and simplifying the constants: -9 + 2 = -7. So vertex form is f(x) = (x + 3)² - 7, with vertex at (-3, -7)! Choice B is correct because it properly completes the square to get vertex form (x + 3)² - 7, revealing the vertex at (-3, -7) with minimum value of -7. Excellent work! Choice C makes a calculation error when finding (b/2)²: half of 6 is 3, and squaring that gives 9, not stopping at -9 without subtracting from the constant. That perfect square term is the key to completing the square, so getting it right matters! The easiest completing the square problems have even middle coefficients: x² + 6x is easier than x² + 7x because half of 6 is 3 (nice and clean), and 3² = 9. Look for these easier patterns when practicing! Remember why we're doing this: vertex form isn't just busywork—it instantly tells you the most important point on the parabola (the vertex) without graphing or calculating. In real-world problems, that vertex often represents the best or worst outcome, like maximum profit or minimum cost. Powerful stuff!
A ball is thrown upward from a height of 6 feet. The height $h(t)$ of the ball after $t$ seconds is given by $h(t) = -16t^2 + 32t + 6$ . What is the maximum height reached by the ball?
Answer: 22 feet To find the maximum height, complete the square: $h(t) = -16t^2 + 32t + 6 = -16(t^2 - 2t) + 6 = -16(t^2 - 2t + 1 - 1) + 6 = -16((t-1)^2 - 1) + 6 = -16(t-1)^2 + 16 + 6 = -16(t-1)^2 + 22$ . The maximum value is 22 feet when $t = 1$ . Choice B incorrectly adds 32 + 6. Choice C uses the coefficient of $t$ as the maximum. Choice D uses half the coefficient of $t^2$ as the maximum.
Complete the square to find the vertex of $f(x)=x^2-8x+3$ . Give your answer as an ordered pair $(h,k)$ .
Answer: $(4,-13)$ This question tests your understanding of completing the square to rewrite a quadratic in vertex form, which reveals the maximum or minimum value. Completing the square transforms a quadratic from standard form (ax² + bx + c) into vertex form (a(x - h)² + k), which immediately shows us the vertex at (h, k)—the highest or lowest point on the parabola—without any calculation needed once we're in that form. Starting with f(x) = x² - 8x + 3, we complete the square on the first two terms: take half of -8 to get -4, square it to get 16, then add and subtract inside: f(x) = (x² - 8x + 16) - 16 + 3. The first part is a perfect square (x - 4)², and simplifying the constants: -16 + 3 = -13. So vertex form is f(x) = (x - 4)² - 13, with vertex at (4, -13)! Choice C is correct because it properly completes the square to get vertex form (x - 4)² - 13, revealing the vertex at (4, -13). Choice A has the right x-coordinate but the wrong y-coordinate—it looks like they added 16 instead of subtracting it when completing the square, getting 3 + 16 = 19 instead of 3 - 16 = -13. Quick memory trick: in vertex form a(x - h)² + k, the signs are tricky! The vertex x-coordinate is h, but in the formula it appears as 'minus h,' so (x - 3)² has vertex at x = 3, while (x + 3)² has vertex at x = -3. The sign flips! But k is straightforward—it's just the constant at the end.
A fireworks shell's height (in feet) after $t$ seconds is modeled by $h(t)=-16t^2+64t+20$ . Complete the square to determine the maximum height of the shell.
Answer: $84$ This question tests your understanding of completing the square to rewrite a quadratic in vertex form, which reveals the maximum or minimum value. To complete the square for x² + bx, we add and subtract (b/2)²—half the middle coefficient, squared. This creates a perfect square trinomial (x + b/2)² and leaves us with extra terms to simplify, giving us the vertex form that shows the vertex clearly. Starting with h(t) = -16t² + 64t + 20, we first factor out -16 from the first two terms: h(t) = -16(t² - 4t) + 20. Inside the parentheses, we complete the square: take half of -4 to get -2, square it to get 4, then add and subtract: h(t) = -16(t² - 4t + 4 - 4) + 20 = -16((t - 2)² - 4) + 20 = -16(t - 2)² + 64 + 20 = -16(t - 2)² + 84. Choice A is correct because from the vertex form h(t) = -16(t - 2)² + 84, we see the vertex at (2, 84), and since the coefficient -16 is negative, the parabola opens downward, making 84 the maximum height. Choice D might tempt you if you forgot to add the original constant 20 to the 64 we got from distributing -16, but the maximum is 64 + 20 = 84 feet. Here's the completing-the-square recipe: (1) Make sure the x² term has coefficient 1 (factor out a if needed), (2) Take half of the x coefficient and square it, (3) Add and subtract that perfect square, (4) Rewrite the first three terms as a perfect square binomial, (5) Simplify the constant terms. Follow these steps and you'll get to vertex form every time!
Rewrite $y=x^2+6x-7$ in vertex form $y=a(x-h)^2+k$ by completing the square.
Answer: $y=(x+3)^2-16$ This question tests your understanding of completing the square to rewrite a quadratic in vertex form, which reveals the maximum or minimum value. To complete the square for x² + bx, we add and subtract (b/2)²—half the middle coefficient, squared. This creates a perfect square trinomial (x + b/2)² and leaves us with extra terms to simplify, giving us the vertex form that shows the vertex clearly. Starting with y = x² + 6x - 7, we complete the square on the first two terms: take half of 6 to get 3, square it to get 9, then add and subtract inside: y = (x² + 6x + 9) - 9 - 7. The first part is a perfect square (x + 3)², and simplifying the constants: -9 - 7 = -16. So vertex form is y = (x + 3)² - 16! Choice A is correct because it properly completes the square to get vertex form (x + 3)² - 16, with the vertex at (-3, -16). Choice B has a sign error in the vertex form: when we have x² + 6x, completing the square gives us (x + 3)², not (x - 3)². Remember that positive 6x means we're adding 3 twice in the perfect square, so it's (x + 3)². The easiest completing the square problems have even middle coefficients: x² + 6x is easier than x² + 7x because half of 6 is 3 (nice and clean), and 3² = 9. Look for these easier patterns when practicing! To check your work: expand your vertex form back out (using FOIL or distributing) and make sure you get back to the original quadratic. If you do, you completed the square correctly!
Find the vertex of the parabola $y=2x^2-8x+1$ by completing the square.
Answer: $(2,-7)$ This question tests your understanding of completing the square to rewrite a quadratic in vertex form, which reveals the maximum or minimum value. Completing the square transforms a quadratic from standard form (ax² + bx + c) into vertex form (a(x - h)² + k), which immediately shows us the vertex at (h, k)—the highest or lowest point on the parabola—without any calculation needed once we're in that form. Starting with y = 2x² - 8x + 1 = 2(x² - 4x) + 1, we complete the square inside: take half of -4 to get -2, square it to get 4, add and subtract: 2(x² - 4x + 4 - 4) + 1 = 2((x - 2)² - 4) + 1 = 2(x - 2)² - 8 + 1. Simplifying the constants: -8 + 1 = -7. So vertex form is y = 2(x - 2)² - 7, with vertex at (2, -7)! Choice A is correct because it properly completes the square to get vertex form 2(x - 2)² - 7, revealing the vertex at (2, -7) with minimum value of -7. Excellent work! Choice B has a sign error in the vertex form: when we write (x - h)², a positive h means (x - positive number), but a negative h means (x - negative number) which looks like (x + positive number). This choice has the sign flipped! In vertex form, the vertex x-coordinate has the opposite sign from what appears in the parentheses. Here's the completing-the-square recipe: (1) Make sure the x² term has coefficient 1 (factor out a if needed), (2) Take half of the x coefficient and square it, (3) Add and subtract that perfect square, (4) Rewrite the first three terms as a perfect square binomial, (5) Simplify the constant terms. Follow these steps and you'll get to vertex form every time! The easiest completing the square problems have even middle coefficients: x² + 6x is easier than x² + 7x because half of 6 is 3 (nice and clean), and 3² = 9. Look for these easier patterns when practicing!
A ball's height (in feet) after $t$ seconds is modeled by $h(t)=-16t^2+32t+5$ . In this model, what is the maximum height of the ball? (Use completing the square.)
Answer: $21$ This question tests your understanding of completing the square to rewrite a quadratic in vertex form, which reveals the maximum or minimum value. The vertex of a parabola is its extreme point: if the parabola opens upward (a > 0), the vertex is the minimum (lowest point), and if it opens downward (a < 0), the vertex is the maximum (highest point). The y-coordinate of the vertex, k, is that extreme value. To find the maximum height in this context, we complete the square on h(t): h(t) = -16(t² - 2t) + 5 = -16(t² - 2t + 1 - 1) + 5 = -16((t - 1)² - 1) + 5 = -16(t - 1)² + 16 + 5 = -16(t - 1)² + 21. From the vertex form, we see the maximum is 21 feet, occurring at t = 1 second. Choice B is correct because it properly completes the square to get vertex form -16(t - 1)² + 21, revealing the vertex at (1, 21) with maximum value of 21. Excellent work! Choice C makes a calculation error when finding (b/2)²: half of -2 (after factoring) is -1, and squaring that gives 1, not something leading to 37. That perfect square term is the key to completing the square, so getting it right matters! Here's the completing-the-square recipe: (1) Make sure the x² term has coefficient 1 (factor out a if needed), (2) Take half of the x coefficient and square it, (3) Add and subtract that perfect square, (4) Rewrite the first three terms as a perfect square binomial, (5) Simplify the constant terms. Follow these steps and you'll get to vertex form every time! To check your work: expand your vertex form back out (using FOIL or distributing) and make sure you get back to the original quadratic. If you do, you completed the square correctly! If not, look for where a sign or number might have slipped.
A ball is thrown upward from a platform. Its height (in feet) after $t$ seconds is modeled by $h(t) = -16t^2 + 64t + 5$ . In context, what is the maximum height of the ball? (Use completing the square to find the vertex.)
Answer: $69$ feet This question tests your understanding of completing the square to rewrite a quadratic in vertex form, which reveals the maximum or minimum value. Completing the square transforms a quadratic from standard form (ax² + bx + c) into vertex form (a(x - h)² + k), which immediately shows us the vertex at (h, k)—the highest or lowest point on the parabola—without any calculation needed once we're in that form. To find the maximum height in this context, we complete the square on h(t) = -16t² + 64t + 5: factor out -16 to get -16(t² - 4t) + 5, half of -4 is -2, square it to 4, add and subtract inside to make -16(t² - 4t + 4 - 4) + 5 = -16[(t - 2)² - 4] + 5 = -16(t - 2)² + 64 + 5 = -16(t - 2)² + 69, so the maximum is 69 feet at 2 seconds. Choice B is correct because it properly completes the square to get vertex form -16(t - 2)² + 69, revealing the vertex at (2, 69) with maximum value of 69. Excellent work! Choice D has a calculation error when finding the constant: after distributing -16 times -4 to get +64, we must add the original +5 to reach 69, not stopping at 64. Here's the completing-the-square recipe: (1) Make sure the x² term has coefficient 1 (factor out a if needed), (2) Take half of the x coefficient and square it, (3) Add and subtract that perfect square, (4) Rewrite the first three terms as a perfect square binomial, (5) Simplify the constant terms. Follow these steps and you'll get to vertex form every time! Remember why we're doing this: vertex form isn't just busywork—it instantly tells you the most important point on the parabola (the vertex) without graphing or calculating. In real-world problems, that vertex often represents the best or worst outcome, like maximum profit or minimum cost. Powerful stuff!
Complete the square to show that $f(x)=-x^2+14x-30$ has a maximum, and find the maximum value.
Answer: $19$ This question tests your understanding of completing the square to rewrite a quadratic in vertex form, which reveals the maximum or minimum value. To complete the square for x² + bx, we add and subtract (b/2)²—half the middle coefficient, squared. This creates a perfect square trinomial (x + b/2)² and leaves us with extra terms to simplify, giving us the vertex form that shows the vertex clearly. Starting with f(x) = -x² + 14x - 30, we factor out -1 from the first two terms: f(x) = -(x² - 14x) - 30. Inside the parentheses, take half of -14 to get -7, square it to get 49: f(x) = -(x² - 14x + 49 - 49) - 30 = -((x - 7)² - 49) - 30 = -(x - 7)² + 49 - 30 = -(x - 7)² + 19. Choice A is correct because from the vertex form f(x) = -(x - 7)² + 19, we see the vertex at (7, 19), and since the coefficient -1 is negative, the parabola opens downward, making 19 the maximum value. Choice B gives the negative of the maximum, perhaps confusing the sign, but the maximum value is positive 19, not -19. Remember why we're doing this: vertex form isn't just busywork—it instantly tells you the most important point on the parabola (the vertex) without graphing or calculating. In real-world problems, that vertex often represents the best or worst outcome, like maximum profit or minimum cost. Powerful stuff!
At what $x$ -value does $f(x) = 2x^2 - 12x + 1$ reach its minimum? (Use completing the square.)
Answer: $x = 3$ This question tests your understanding of completing the square to rewrite a quadratic in vertex form, which reveals the maximum or minimum value. The vertex of a parabola is its extreme point: if the parabola opens upward (a > 0), the vertex is the minimum (lowest point), and if it opens downward (a < 0), the vertex is the maximum (highest point). The y-coordinate of the vertex, k, is that extreme value. Starting with f(x) = 2x² - 12x + 1, factor out 2 from the x terms: 2(x² - 6x) + 1, take half of -6 to get -3, square to 9, add and subtract inside: 2(x² - 6x + 9 - 9) + 1 = 2[(x - 3)² - 9] + 1 = 2(x - 3)² - 18 + 1 = 2(x - 3)² - 17, so the vertex is at (3, -17). Choice C is correct because it properly completes the square to get vertex form 2(x - 3)² - 17, revealing the vertex at (3, -17) with minimum occurring at x=3. Excellent work! Choice A has a sign error in the vertex form: when we write (x - h)², a positive h means (x - positive number), but a negative h means (x - negative number) which looks like (x + positive number). This choice has the sign flipped! In vertex form, the vertex x-coordinate has the opposite sign from what appears in the parentheses. Quick memory trick: in vertex form a(x - h)² + k, the signs are tricky! The vertex x-coordinate is h, but in the formula it appears as 'minus h,' so (x - 3)² has vertex at x = 3, while (x + 3)² has vertex at x = -3. The sign flips! But k is straightforward—it's just the constant at the end.
Rewrite $y = -x^2 + 10x - 7$ in vertex form $y = a(x - h)^2 + k$ .
Answer: $y = -(x - 5)^2 + 18$ This question tests your understanding of completing the square to rewrite a quadratic in vertex form, which reveals the maximum or minimum value. Completing the square transforms a quadratic from standard form (ax² + bx + c) into vertex form (a(x - h)² + k), which immediately shows us the vertex at (h, k)—the highest or lowest point on the parabola—without any calculation needed once we're in that form. Starting with y = -x² + 10x - 7, factor out -1 from the x terms: y = - (x² - 10x) - 7, take half of -10 to get -5, square to 25, add and subtract inside: - (x² - 10x + 25 - 25) - 7 = - [(x - 5)² - 25] - 7 = - (x - 5)² + 25 - 7 = - (x - 5)² + 18. Choice B is correct because it properly completes the square to get vertex form -(x - 5)² + 18, revealing the vertex at (5, 18) with maximum value of 18. Excellent work! Choice A has a sign error in the vertex form: when we write (x - h)², a positive h means (x - positive number), but a negative h means (x - negative number) which looks like (x + positive number). This choice has the sign flipped! In vertex form, the vertex x-coordinate has the opposite sign from what appears in the parentheses. Here's the completing-the-square recipe: (1) Make sure the x² term has coefficient 1 (factor out a if needed), (2) Take half of the x coefficient and square it, (3) Add and subtract that perfect square, (4) Rewrite the first three terms as a perfect square binomial, (5) Simplify the constant terms. Follow these steps and you'll get to vertex form every time! To check your work: expand your vertex form back out (using FOIL or distributing) and make sure you get back to the original quadratic. If you do, you completed the square correctly! If not, look for where a sign or number might have slipped.
Rewrite $f(x)=2x^2-8x+1$ in vertex form $a(x-h)^2+k$ by completing the square.
Answer: $f(x)=2(x-2)^2-7$ This question tests your understanding of completing the square to rewrite a quadratic in vertex form, which reveals the maximum or minimum value. Completing the square transforms a quadratic from standard form (ax² + bx + c) into vertex form (a(x - h)² + k), which immediately shows us the vertex at (h, k)—the highest or lowest point on the parabola—without any calculation needed once we're in that form. Starting with f(x) = 2x² - 8x + 1, we first factor out 2 from the first two terms: f(x) = 2(x² - 4x) + 1. Inside the parentheses, we complete the square: take half of -4 to get -2, square it to get 4, then add and subtract: f(x) = 2(x² - 4x + 4 - 4) + 1 = 2((x - 2)² - 4) + 1 = 2(x - 2)² - 8 + 1 = 2(x - 2)² - 7. Choice B is correct because it properly completes the square to get vertex form 2(x - 2)² - 7, revealing the vertex at (2, -7). Choice C makes a calculation error with the constant term: when we distribute 2(-4) we get -8, and -8 + 1 = -7, not +9. The easiest completing the square problems have even middle coefficients: x² + 6x is easier than x² + 7x because half of 6 is 3 (nice and clean), and 3² = 9. Look for these easier patterns when practicing! To check your work: expand your vertex form back out (using FOIL or distributing) and make sure you get back to the original quadratic.

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