Graphing Quadratic Functions and Conic Sections - SAT Math

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Question

What is the center and radius of the circle indicated by the equation?

Answer

A circle is defined by an equation in the format .

The center is indicated by the point and the radius .

In the equation , the center is and the radius is .

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Question

Give the -coordinate of the vertex of the parabola of the function

$f(x) = 2x^{2} + 6x + 5$ .

Answer

The -coordinate of the vertex of a parabola of the form

$f(x) = ax^{2}+ bx + c$

is

$x= -\frac{b}{2a}$ .

Set :

$x= -\frac{6}{2\cdot 2} = -\frac{6}{4} =- \frac{3}{2}$

The -coordinate is therefore $f\left (- \frac{3}{2} \right )$ :

$f(x) = 2x^{2} + 6x + 5$

$f \left ( -\frac{3}{2} \right )= 2\left ( -\frac{3}{2} \right )^{2}+ 6\left ( -\frac{3}{2} \right ) + 5$

$= 2\left ( \frac{9}{4} \right ) + 6\left (- \frac{3}{2} \right ) + 5$

$= \frac{9}{2} - 9 + 5 = \frac{1}{2}$ , which is the correct choice.

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Question

Give the axis of symmetry of the parabola of the equation

$f(x) = 3x^{2} + 9x - 16$

Answer

The line of symmetry of the parabola of the equation

$f(x) = ax^{2}+ bx + c$

is the vertical line

$x = -\frac{b}{2a}$

Substitute :

The line of symmetry is

$x = -\frac{9}{2 \cdot 3} = - \frac{9}{6} = -\frac{3}{2}$

That is, the line of the equation $x =- \frac{3}{2}$ .

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Question

Give the -intercept(s) of the parabola of the equation

$f(x) = 3x^{2}-9x -54$

Answer

Set and solve for :

$f(x) = 3x^{2}-9x -54$

$3x^{2}-9x -54 = 0$

The terms have a GCF of 2, so

$3\left ( x^{2}- 3x - 18 \right )= 0$

The trinomial in parentheses can be FOILed out by noting that and $-6 \times 3= -18$ :

And you can divide both sides by 3 to get rid of the coefficient:

Set each of the linear binomials to 0 and solve for :

$x - 6 = 0 \Rightarrow x = 6$

or

$x +3= 0 \Rightarrow x = -3$

The parabola has as its two intercepts the points and .

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Question

Give the -coordinate of the vertex of the parabola of the function

$f(x) = 4x^{2}- 16x + 1$

Answer

The -coordinate of the vertex of a parabola of the form

$f(x) = ax^{2}+ bx + c$

is

$x= -\frac{b}{2a}$ .

Substitute :

$x= -\frac{-16}{2\cdot 4} = -\frac{-16}{8} = 2$

The -coordinate is therefore :

$f(x) = 4x^{2}- 16x + 1$

$f(2) = 4 \cdot 2 ^{2}- 16 \cdot 2 + 1$

$f(2) = 4 \cdot 4- 16 \cdot 2 + 1$

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Question

A baseball is thrown straight up with an initial speed of 60 miles per hour by a man standing on the roof of a 100-foot high building. The height of the baseball in feet is modeled by the function

$h(t) = -16t^{2}+60t + 100$

To the nearest foot, how high is the baseball when it reaches the highest point of its path?

Answer

We are seeking the value of when the graph of $h\left ( t\right )$ - a parabola - reaches its vertex.

To find this value, we first find the value of . For a parabola of the equation

$f\left ( t\right ) = at^{2}+ bt + c$ ,

the value of the vertex is

$t = -\frac{b}{2a}$ .

Substitute :

$t = -\frac{60}{2 (-16)} = 1.875$

The height of the baseball after 1.875 seconds will be

$h(1.875) = -16 \cdot 1.875^{2}+60 \cdot 1.875 + 100$

$h(1.875) = -56.25 +112.5 + 100 \approx 156$ feet.

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Question

A baseball is thrown upward from the top of a one hundred and fifty-foot-high building. The initial speed of the ball is forty-five miles per hour. The height of the ball after seconds is modeled by the function

$h (t) = -16t^{2}+ 45t +150$

How high does the ball get (nearest foot)?

Answer

A quadratic function such as has a parabola as its graph. The high point of the parabola - the vertex - is what we are looking for.

The vertex of a function

$h (t) = at^{2}+bt +c$

has as coordinates

$\left ( - \frac{b}{2a} , h \left ( - \frac{b}{2a} \right ) \right )$ .

The second coordinate is the height and we are looking for this quantity. Since $h (t) = -16t^{2}+ 45t +150$ , setting :

$- \frac{b}{2a} = - \frac{45}{2(-16)} = \frac{45}{32} = 1.40625$ seconds for the ball to reach the peak.

The height of the ball at this point is , which can be evaluated by substitution:

$h (t) = -16t^{2}+ 45t +150$

$h (1.40625 ) = -16 (1.40625 )^{2}+ 45 (1.40625 ) +150$

$\approx -16 (1.9775 ) + 45 (1.40625 ) +150$

$\approx -31.6406 + 63.2813 + 150$

$\approx 181.6407$

Round this to 182 feet.

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Question

What is the center and radius of the circle indicated by the equation?

Answer

A circle is defined by an equation in the format .

The center is indicated by the point and the radius .

In the equation , the center is and the radius is .

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Question

Give the -coordinate of the vertex of the parabola of the function

$f(x) = 2x^{2} + 6x + 5$ .

Answer

The -coordinate of the vertex of a parabola of the form

$f(x) = ax^{2}+ bx + c$

is

$x= -\frac{b}{2a}$ .

Set :

$x= -\frac{6}{2\cdot 2} = -\frac{6}{4} =- \frac{3}{2}$

The -coordinate is therefore $f\left (- \frac{3}{2} \right )$ :

$f(x) = 2x^{2} + 6x + 5$

$f \left ( -\frac{3}{2} \right )= 2\left ( -\frac{3}{2} \right )^{2}+ 6\left ( -\frac{3}{2} \right ) + 5$

$= 2\left ( \frac{9}{4} \right ) + 6\left (- \frac{3}{2} \right ) + 5$

$= \frac{9}{2} - 9 + 5 = \frac{1}{2}$ , which is the correct choice.

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Question

Give the axis of symmetry of the parabola of the equation

$f(x) = 3x^{2} + 9x - 16$

Answer

The line of symmetry of the parabola of the equation

$f(x) = ax^{2}+ bx + c$

is the vertical line

$x = -\frac{b}{2a}$

Substitute :

The line of symmetry is

$x = -\frac{9}{2 \cdot 3} = - \frac{9}{6} = -\frac{3}{2}$

That is, the line of the equation $x =- \frac{3}{2}$ .

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Question

Give the -intercept(s) of the parabola of the equation

$f(x) = 3x^{2}-9x -54$

Answer

Set and solve for :

$f(x) = 3x^{2}-9x -54$

$3x^{2}-9x -54 = 0$

The terms have a GCF of 2, so

$3\left ( x^{2}- 3x - 18 \right )= 0$

The trinomial in parentheses can be FOILed out by noting that and $-6 \times 3= -18$ :

And you can divide both sides by 3 to get rid of the coefficient:

Set each of the linear binomials to 0 and solve for :

$x - 6 = 0 \Rightarrow x = 6$

or

$x +3= 0 \Rightarrow x = -3$

The parabola has as its two intercepts the points and .

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Question

Give the -coordinate of the vertex of the parabola of the function

$f(x) = 4x^{2}- 16x + 1$

Answer

The -coordinate of the vertex of a parabola of the form

$f(x) = ax^{2}+ bx + c$

is

$x= -\frac{b}{2a}$ .

Substitute :

$x= -\frac{-16}{2\cdot 4} = -\frac{-16}{8} = 2$

The -coordinate is therefore :

$f(x) = 4x^{2}- 16x + 1$

$f(2) = 4 \cdot 2 ^{2}- 16 \cdot 2 + 1$

$f(2) = 4 \cdot 4- 16 \cdot 2 + 1$

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Question

A baseball is thrown straight up with an initial speed of 60 miles per hour by a man standing on the roof of a 100-foot high building. The height of the baseball in feet is modeled by the function

$h(t) = -16t^{2}+60t + 100$

To the nearest foot, how high is the baseball when it reaches the highest point of its path?

Answer

We are seeking the value of when the graph of $h\left ( t\right )$ - a parabola - reaches its vertex.

To find this value, we first find the value of . For a parabola of the equation

$f\left ( t\right ) = at^{2}+ bt + c$ ,

the value of the vertex is

$t = -\frac{b}{2a}$ .

Substitute :

$t = -\frac{60}{2 (-16)} = 1.875$

The height of the baseball after 1.875 seconds will be

$h(1.875) = -16 \cdot 1.875^{2}+60 \cdot 1.875 + 100$

$h(1.875) = -56.25 +112.5 + 100 \approx 156$ feet.

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Question

A baseball is thrown upward from the top of a one hundred and fifty-foot-high building. The initial speed of the ball is forty-five miles per hour. The height of the ball after seconds is modeled by the function

$h (t) = -16t^{2}+ 45t +150$

How high does the ball get (nearest foot)?

Answer

A quadratic function such as has a parabola as its graph. The high point of the parabola - the vertex - is what we are looking for.

The vertex of a function

$h (t) = at^{2}+bt +c$

has as coordinates

$\left ( - \frac{b}{2a} , h \left ( - \frac{b}{2a} \right ) \right )$ .

The second coordinate is the height and we are looking for this quantity. Since $h (t) = -16t^{2}+ 45t +150$ , setting :

$- \frac{b}{2a} = - \frac{45}{2(-16)} = \frac{45}{32} = 1.40625$ seconds for the ball to reach the peak.

The height of the ball at this point is , which can be evaluated by substitution:

$h (t) = -16t^{2}+ 45t +150$

$h (1.40625 ) = -16 (1.40625 )^{2}+ 45 (1.40625 ) +150$

$\approx -16 (1.9775 ) + 45 (1.40625 ) +150$

$\approx -31.6406 + 63.2813 + 150$

$\approx 181.6407$

Round this to 182 feet.

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Question

What is the center and radius of the circle indicated by the equation?

Answer

A circle is defined by an equation in the format .

The center is indicated by the point and the radius .

In the equation , the center is and the radius is .

Compare your answer with the correct one above

Question

Give the -coordinate of the vertex of the parabola of the function

$f(x) = 2x^{2} + 6x + 5$ .

Answer

The -coordinate of the vertex of a parabola of the form

$f(x) = ax^{2}+ bx + c$

is

$x= -\frac{b}{2a}$ .

Set :

$x= -\frac{6}{2\cdot 2} = -\frac{6}{4} =- \frac{3}{2}$

The -coordinate is therefore $f\left (- \frac{3}{2} \right )$ :

$f(x) = 2x^{2} + 6x + 5$

$f \left ( -\frac{3}{2} \right )= 2\left ( -\frac{3}{2} \right )^{2}+ 6\left ( -\frac{3}{2} \right ) + 5$

$= 2\left ( \frac{9}{4} \right ) + 6\left (- \frac{3}{2} \right ) + 5$

$= \frac{9}{2} - 9 + 5 = \frac{1}{2}$ , which is the correct choice.

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Question

Give the axis of symmetry of the parabola of the equation

$f(x) = 3x^{2} + 9x - 16$

Answer

The line of symmetry of the parabola of the equation

$f(x) = ax^{2}+ bx + c$

is the vertical line

$x = -\frac{b}{2a}$

Substitute :

The line of symmetry is

$x = -\frac{9}{2 \cdot 3} = - \frac{9}{6} = -\frac{3}{2}$

That is, the line of the equation $x =- \frac{3}{2}$ .

Compare your answer with the correct one above

Question

Give the -intercept(s) of the parabola of the equation

$f(x) = 3x^{2}-9x -54$

Answer

Set and solve for :

$f(x) = 3x^{2}-9x -54$

$3x^{2}-9x -54 = 0$

The terms have a GCF of 2, so

$3\left ( x^{2}- 3x - 18 \right )= 0$

The trinomial in parentheses can be FOILed out by noting that and $-6 \times 3= -18$ :

And you can divide both sides by 3 to get rid of the coefficient:

Set each of the linear binomials to 0 and solve for :

$x - 6 = 0 \Rightarrow x = 6$

or

$x +3= 0 \Rightarrow x = -3$

The parabola has as its two intercepts the points and .

Compare your answer with the correct one above

Question

Give the -coordinate of the vertex of the parabola of the function

$f(x) = 4x^{2}- 16x + 1$

Answer

The -coordinate of the vertex of a parabola of the form

$f(x) = ax^{2}+ bx + c$

is

$x= -\frac{b}{2a}$ .

Substitute :

$x= -\frac{-16}{2\cdot 4} = -\frac{-16}{8} = 2$

The -coordinate is therefore :

$f(x) = 4x^{2}- 16x + 1$

$f(2) = 4 \cdot 2 ^{2}- 16 \cdot 2 + 1$

$f(2) = 4 \cdot 4- 16 \cdot 2 + 1$

Compare your answer with the correct one above

Question

A baseball is thrown straight up with an initial speed of 60 miles per hour by a man standing on the roof of a 100-foot high building. The height of the baseball in feet is modeled by the function

$h(t) = -16t^{2}+60t + 100$

To the nearest foot, how high is the baseball when it reaches the highest point of its path?

Answer

We are seeking the value of when the graph of $h\left ( t\right )$ - a parabola - reaches its vertex.

To find this value, we first find the value of . For a parabola of the equation

$f\left ( t\right ) = at^{2}+ bt + c$ ,

the value of the vertex is

$t = -\frac{b}{2a}$ .

Substitute :

$t = -\frac{60}{2 (-16)} = 1.875$

The height of the baseball after 1.875 seconds will be

$h(1.875) = -16 \cdot 1.875^{2}+60 \cdot 1.875 + 100$

$h(1.875) = -56.25 +112.5 + 100 \approx 156$ feet.

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