This question asks for the factored form of (y = $x^2$ + 8x + 7). Factor by finding numbers that multiply to 7 and add to 8, which are 1 and 7, so (y = (x + 1)(x + 7)). Verify by expanding: $(x^2$ + 7x + x + 7 = $x^2$ + 8x + 7). The quadratic formula gives roots -1 and -7, confirming factors. Avoid errors like choosing numbers that multiply incorrectly, such as -1 and -7 which add to -8. When factoring quadratics with positive constant and leading coefficient 1, look for positive factors if b>0.

This question asks us to find the solutions to the quadratic equation $2x^2-5x-3=0$. We can solve this by factoring: we need two numbers that multiply to $(2)(-3)=-6$ and add to $-5$. These numbers are $-6$ and $1$, so we rewrite as $2x^2-6x+x-3=0$, then factor by grouping: $2x(x-3)+1(x-3)=0$, giving us $(2x+1)(x-3)=0$. Setting each factor to zero: $2x+1=0$ gives $x=-\frac{1}{2}$, and $x-3=0$ gives $x=3$. A common error is making arithmetic mistakes when factoring or solving for x. When you see a quadratic equation on the PSAT, first check if it factors nicely before using the quadratic formula.

This question asks for the time when the ball reaches maximum height, given the height function $h(t)=-16t^2+48t+5$. For a quadratic function in the form $at^2+bt+c$, the maximum (when $a<0$) occurs at $t=-\frac{b}{2a}$. Here, $a=-16$ and $b=48$, so the maximum occurs at $t=-\frac{48}{2(-16)}=-\frac{48}{-32}=\frac{48}{32}=\frac{3}{2}$. We can verify this is a maximum since the coefficient of $t^2$ is negative, meaning the parabola opens downward. A common error is forgetting the negative sign in the formula or making arithmetic mistakes with fractions. For projectile motion problems, remember that the vertex formula gives you the time of maximum height directly.

The axis of symmetry of a parabola is a vertical line that passes through the vertex. For the parabola $y=(x+1)^2-4$ in vertex form, the vertex is at $(-1,-4)$ because we have $(x-h)$ where $h=-1$. The axis of symmetry is the vertical line through this vertex, which has equation $x=-1$. Students often confuse the axis of symmetry (a vertical line $x=$ constant) with horizontal lines ($y=$ constant). Remember that parabolas with vertical axes have vertical lines of symmetry, making the answer format $x=$ value.

The question asks for the axis of symmetry of y = -2x² + 8x - 3. The axis is the vertical line x = -b/(2a), with a = -2, b = 8, so x = -8/(2*(-2)) = -8/-4 = 2. This is the line through the vertex, confirmed by vertex form after completing the square: y = -2(x² - 4x) - 3 = -2(x² - 4x + 4 - 4) - 3 = -2(x - 2)² + 8 - 3 = -2(x - 2)² + 5. Mistaking it for y-values like the y-intercept -3 or another number is common. Confusing with positive b sign can lead to x = -2. Always use the formula x = -b/(2a) for quick identification in graphing questions.

The question asks when the ball reaches maximum height given $h(t)=-5t^2+20t+1$. For a quadratic in the form $at^2+bt+c$, the maximum (when $a<0$) occurs at $t = -\frac{b}{2a}$. Here, $a=-5$ and $b=20$, so $t = -\frac{20}{2(-5)} = -\frac{20}{-10} = 2$ seconds. We can verify this is a maximum since the coefficient of $t^2$ is negative (-5), making the parabola open downward. A common mistake is forgetting the negative sign in the formula or confusing when to use maximum vs minimum. For time-based problems, always check that your answer makes physical sense (positive time).

This question seeks the time (t) when a ball reaches maximum height, modeled by (h(t) = $-5t^2$ + 20t + 1). For quadratics, the vertex gives the maximum (since a<0), at (t = -rac{b}{2a} = -rac{20}{2(-5)} = rac{20}{10} = 2). Completing the square: (h(t) = $-5(t^2$ - 4t) + 1 = -5(t - $2)^2$ + 21), confirming peak at t=2. The quadratic formula isn't needed for vertex time. A common mistake is using positive a or forgetting the negative sign. In projectile problems, remember the vertex formula directly gives the time to max height.

This question asks to solve $(2x^2$ + 3x - 2 = 0) and select the solutions. Factor: find numbers for (2x + a)(x + b) where a*b=-2, and cross terms give 3x; try (2x - 1)(x + 2) = $2x^2$ + 4x - x - 2 = $2x^2$ + 3x - 2, yes, so x=1/2 or x=-2. Quadratic formula: (x = rac{-3 pm $\sqrt{9 + 16}$}{4} = rac{-3 pm 5}{4}), giving (rac{2}{4}=0.5) and (rac{-8}{4}=-2). Factoring works well with small integers; formula confirms. Common error: incorrect factoring signs. Verify by plugging solutions back in.

The vertex form $y=a(x-h)^2+k$ directly shows the vertex at $(h,k)$. In the equation $y=2(x-3)^2-8$, we identify $h=3$ and $k=-8$, so the vertex is at $(3,-8)$. Note that the sign inside the parentheses is opposite to the x-coordinate: $(x-3)$ means the vertex has x-coordinate positive 3. A common error is thinking $(x-3)$ gives vertex at $x=-3$, but remember the form is $(x-h)$, not $(x+h)$. When given vertex form, you can immediately read off the vertex without any calculations.

To convert $y=x^2+6x+1$ to vertex form, we complete the square. First, focus on $x^2+6x$: we take half of 6 (which is 3) and square it (getting 9). So $x^2+6x = (x+3)^2 - 9$. Therefore, $y = x^2+6x+1 = (x+3)^2 - 9 + 1 = (x+3)^2 - 8$. The vertex form is $y=(x+3)^2-8$, showing the vertex at $(-3,-8)$. A common error is forgetting to subtract the 9 when completing the square, or making sign errors. Always expand your final answer to verify it matches the original equation.