How to use the quadratic function - PSAT Math

Card 0 of 63

If x2 + 2x - 1 = 7, which answers for x are correct?

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x2 + 2x - 1 = 7

x2 + 2x - 8 = 0

(x + 4) (x - 2) = 0

x = -4, x = 2

Which of the following quadratic equations has a vertex located at (3,4)?

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The vertex form of a parabola is given by the equation:

$f(x)=a(x-h)^2$ +k, where the point (h,k) is the vertex, and a is a constant.

We are told that the vertex must occur at (3,4), so let's plug this information into the vertex form of the equation. h will be 3, and k will be 4.

$f(x)=a(x-3)^2$ +4

Let's now expand $(x-3)^2$ by using the FOIL method, which requires us to multiply the first, inner, outer, and last terms together before adding them all together.

$(x-3)^2$ = $(x-3)(x-3)=x^2$$-3x-3x+9=x^2$-6x+9

We can replace $(x-3)^2$ with $x^2$-6x+9.

$f(x)=a(x-3)^2$$+4=a(x^2$-6x+9)+4

Next, distribute the a.

$a(x^2$-6x+9)+4 = $ax^2$ -6ax+9a+4

Notice that in all of our answer choices, the first term is $-2x^2$. If we let a=-2, then we would have $-2x^2$ in our equation. Let's see what happens when we substitute -2 for a.

$f(x)=ax^2$$-6ax+9a+4=(-2)x^2$-6(-2)x+9(-2)+4

$=-2x^2$+12x-18+4

Use the quadratic equation to solve for $\small x$ .

$\small \small \small \small $x^{2}$ + 3x - 3 = 0$

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We take a polynomial in the form $\small $ax^2$ + bx +c = 0$

and enter the corresponding coefficients into the quadratic equation.

$\small \small $\frac{-b\pm $\sqrt{b^2$ - 4ac}$}{2a} = x$ . We normally expect to have two answers given by the sign $\small \pm$ .

So,

$\small \small \small \small $x^{2}$ + 3x - 3 = 0$

$\small \small \small \small $\frac{-(3)\pm $\sqrt{(3)^2$ - 4(1)(-3)}$}{2(1)} = x$

$\small \small \small \small $\frac{-3\pm \sqrt{9 + 12}$}{2} = x$

$\small \small \small \small \small $\frac{-3\pm \sqrt{21}$}{2} = x$

Define function as follows:

$f(x) = $x^{2}$ - 10x + 7$

Given that and $a \ne 5$ , evaluate .

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$f(x) = $x^{2}$ - 10x + 7$

$f(5) = $5^{2}$ - 10 \cdot 5 + 7$

We solve for in the equation

This is the only solution. Since it is established that is not equal to 5, the correct response is that no such value exists.

A pitcher standing on top on a 600-foot-high building throws a baseball upward at an initial speed of 90 feet per second. The height of the ball at a given time can be modeled by the function

$h\left ( t\right )= $-16t^{2}$+ 90 t + 600$

How long does it take for the ball to hit the ground? (Nearest tenth of a second)

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Set the height function equal to 0:

$h\left ( t\right )=0$

Set in the quadratic formula

$t = $\frac{-b \pm $\sqrt{b^{2}$$-4ac}}{2a}$

$t = $\frac{-90 \pm $\sqrt{90^{2}$$-4(-16)(600)}}{2 (-16)}$

$t = $\frac{-90 \pm \sqrt{8,100+38,400}$}{-32}$

$t = $\frac{-90 \pm \sqrt{46,500}$}{-32}$

$t \approx $\frac{-90 \pm 215.64}{-32}$$

Evaluate separately, and select the positive value.

$t \approx $\frac{-90 +215.64}{-32}$ \approx $\frac{125.64}{-32}$ \approx -3.9$

which is thrown out.

$t \approx $\frac{-90 -215.64}{-32}$ \approx $\frac{-305.64}{-32}$ \approx 9.6$

which is positive and is the result we keep.

The correct response is 9.6 seconds.

A pitcher standing on top on a 600-foot building throws a baseball upward at an initial speed of 90 feet per second. The height of the ball at a given time can be modeled by the function

$h\left ( t\right )= $-16t^{2}$+ 90 t + 600$

How long does it take for the ball to reach its peak? (Nearest tenth of a second)

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The time at which the ball reaches its peak can be found by finding the -coordinate of the vertex of the parabola representing the function.

The -coordinate of the vertex is $-$\frac{b}{2a}$$ , where ;

$-$\frac{b}{2a}$ = -$\frac{90}{2(-16) }$ = 2.8$

The correct response is 2.8 seconds.

A pitcher standing on top on a 600-foot building throws a baseball upward at an initial speed of 90 feet per second. The height of the ball at a given time can be modeled by the function

$h\left ( t\right )= $-16t^{2}$+ 90 t + 600$

How high does the ball get? (Nearest foot)

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This can be solved by first finding the first coordinate ( value) of the vertex of the parabola representing the function.

The -coordinate of the vertex is $-$\frac{b}{2a}$$ , where ;

$-$\frac{b}{2a}$ = -$\frac{90}{2(-16) }$ = 2.8$

The ball takes 2.8 seconds to reach its peak. The height at that time is , which is evaluated using substitution:

$h\left ( t\right )= $-16t^{2}$+ 90 t + 600$

$h\left ( 2.8 \right )= -16\cdot $2.8^{2}$+ 90 \cdot 2.8+ 600$

$= -16\cdot 7.84+ 90 \cdot 2.8+ 600$

making the correct response 727 feet.

Define function as follows:

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

Given that and $a \ne 9$ , evaluate .

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

$f(9) = $9^{2}$ - 6 \cdot 9 + 5$

Solve for in the equation

Either , in which case , which is already established to be untrue, or , in which case . This is the correct response.

A pitcher standing on top on a 600-foot building throws a baseball upward at an initial speed of 90 feet per second. The height of the ball at a given time can be modeled by the function

$h\left ( t\right )= $-16t^{2}$+ 90 t + 600$

How long does it take for the ball to return to the level at which it started? (Nearest tenth of a second)

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The question is essentially asking for the positive value of time when

.

$-2 t \left (8t - 45 \right ) = 0$

Either - but this simply reflects that the initial height was 600 feet - or:

$t = 45 \div 8 \approx 5.6$ .

The ball returns to a height of 600 feet after 5.6 seconds.

If x2 + 2x - 1 = 7, which answers for x are correct?

Tap to see back →

x2 + 2x - 1 = 7

x2 + 2x - 8 = 0

(x + 4) (x - 2) = 0

x = -4, x = 2

Which of the following quadratic equations has a vertex located at (3,4)?

Tap to see back →

The vertex form of a parabola is given by the equation:

$f(x)=a(x-h)^2$ +k, where the point (h,k) is the vertex, and a is a constant.

We are told that the vertex must occur at (3,4), so let's plug this information into the vertex form of the equation. h will be 3, and k will be 4.

$f(x)=a(x-3)^2$ +4

Let's now expand $(x-3)^2$ by using the FOIL method, which requires us to multiply the first, inner, outer, and last terms together before adding them all together.

$(x-3)^2$ = $(x-3)(x-3)=x^2$$-3x-3x+9=x^2$-6x+9

We can replace $(x-3)^2$ with $x^2$-6x+9.

$f(x)=a(x-3)^2$$+4=a(x^2$-6x+9)+4

Next, distribute the a.

$a(x^2$-6x+9)+4 = $ax^2$ -6ax+9a+4

Notice that in all of our answer choices, the first term is $-2x^2$. If we let a=-2, then we would have $-2x^2$ in our equation. Let's see what happens when we substitute -2 for a.

$f(x)=ax^2$$-6ax+9a+4=(-2)x^2$-6(-2)x+9(-2)+4

$=-2x^2$+12x-18+4

Use the quadratic equation to solve for $\small x$ .

$\small \small \small \small $x^{2}$ + 3x - 3 = 0$

Tap to see back →

We take a polynomial in the form $\small $ax^2$ + bx +c = 0$

and enter the corresponding coefficients into the quadratic equation.

$\small \small $\frac{-b\pm $\sqrt{b^2$ - 4ac}$}{2a} = x$ . We normally expect to have two answers given by the sign $\small \pm$ .

So,

$\small \small \small \small $x^{2}$ + 3x - 3 = 0$

$\small \small \small \small $\frac{-(3)\pm $\sqrt{(3)^2$ - 4(1)(-3)}$}{2(1)} = x$

$\small \small \small \small $\frac{-3\pm \sqrt{9 + 12}$}{2} = x$

$\small \small \small \small \small $\frac{-3\pm \sqrt{21}$}{2} = x$

Define function as follows:

$f(x) = $x^{2}$ - 10x + 7$

Given that and $a \ne 5$ , evaluate .

Tap to see back →

$f(x) = $x^{2}$ - 10x + 7$

$f(5) = $5^{2}$ - 10 \cdot 5 + 7$

We solve for in the equation

This is the only solution. Since it is established that is not equal to 5, the correct response is that no such value exists.

A pitcher standing on top on a 600-foot-high building throws a baseball upward at an initial speed of 90 feet per second. The height of the ball at a given time can be modeled by the function

$h\left ( t\right )= $-16t^{2}$+ 90 t + 600$

How long does it take for the ball to hit the ground? (Nearest tenth of a second)

Tap to see back →

Set the height function equal to 0:

$h\left ( t\right )=0$

Set in the quadratic formula

$t = $\frac{-b \pm $\sqrt{b^{2}$$-4ac}}{2a}$

$t = $\frac{-90 \pm $\sqrt{90^{2}$$-4(-16)(600)}}{2 (-16)}$

$t = $\frac{-90 \pm \sqrt{8,100+38,400}$}{-32}$

$t = $\frac{-90 \pm \sqrt{46,500}$}{-32}$

$t \approx $\frac{-90 \pm 215.64}{-32}$$

Evaluate separately, and select the positive value.

$t \approx $\frac{-90 +215.64}{-32}$ \approx $\frac{125.64}{-32}$ \approx -3.9$

which is thrown out.

$t \approx $\frac{-90 -215.64}{-32}$ \approx $\frac{-305.64}{-32}$ \approx 9.6$

which is positive and is the result we keep.

The correct response is 9.6 seconds.

A pitcher standing on top on a 600-foot building throws a baseball upward at an initial speed of 90 feet per second. The height of the ball at a given time can be modeled by the function

$h\left ( t\right )= $-16t^{2}$+ 90 t + 600$

How long does it take for the ball to reach its peak? (Nearest tenth of a second)

Tap to see back →

The time at which the ball reaches its peak can be found by finding the -coordinate of the vertex of the parabola representing the function.

The -coordinate of the vertex is $-$\frac{b}{2a}$$ , where ;

$-$\frac{b}{2a}$ = -$\frac{90}{2(-16) }$ = 2.8$

The correct response is 2.8 seconds.

A pitcher standing on top on a 600-foot building throws a baseball upward at an initial speed of 90 feet per second. The height of the ball at a given time can be modeled by the function

$h\left ( t\right )= $-16t^{2}$+ 90 t + 600$

How high does the ball get? (Nearest foot)

Tap to see back →

This can be solved by first finding the first coordinate ( value) of the vertex of the parabola representing the function.

The -coordinate of the vertex is $-$\frac{b}{2a}$$ , where ;

$-$\frac{b}{2a}$ = -$\frac{90}{2(-16) }$ = 2.8$

The ball takes 2.8 seconds to reach its peak. The height at that time is , which is evaluated using substitution:

$h\left ( t\right )= $-16t^{2}$+ 90 t + 600$

$h\left ( 2.8 \right )= -16\cdot $2.8^{2}$+ 90 \cdot 2.8+ 600$

$= -16\cdot 7.84+ 90 \cdot 2.8+ 600$

making the correct response 727 feet.

Define function as follows:

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

Given that and $a \ne 9$ , evaluate .

Tap to see back →

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

$f(9) = $9^{2}$ - 6 \cdot 9 + 5$

Solve for in the equation

Either , in which case , which is already established to be untrue, or , in which case . This is the correct response.

A pitcher standing on top on a 600-foot building throws a baseball upward at an initial speed of 90 feet per second. The height of the ball at a given time can be modeled by the function

$h\left ( t\right )= $-16t^{2}$+ 90 t + 600$

How long does it take for the ball to return to the level at which it started? (Nearest tenth of a second)

Tap to see back →

The question is essentially asking for the positive value of time when

.

$-2 t \left (8t - 45 \right ) = 0$

Either - but this simply reflects that the initial height was 600 feet - or:

$t = 45 \div 8 \approx 5.6$ .

The ball returns to a height of 600 feet after 5.6 seconds.

If x2 + 2x - 1 = 7, which answers for x are correct?

Tap to see back →

x2 + 2x - 1 = 7

x2 + 2x - 8 = 0

(x + 4) (x - 2) = 0

x = -4, x = 2

Which of the following quadratic equations has a vertex located at (3,4)?

Tap to see back →

The vertex form of a parabola is given by the equation:

$f(x)=a(x-h)^2$ +k, where the point (h,k) is the vertex, and a is a constant.

We are told that the vertex must occur at (3,4), so let's plug this information into the vertex form of the equation. h will be 3, and k will be 4.

$f(x)=a(x-3)^2$ +4

Let's now expand $(x-3)^2$ by using the FOIL method, which requires us to multiply the first, inner, outer, and last terms together before adding them all together.

$(x-3)^2$ = $(x-3)(x-3)=x^2$$-3x-3x+9=x^2$-6x+9

We can replace $(x-3)^2$ with $x^2$-6x+9.

$f(x)=a(x-3)^2$$+4=a(x^2$-6x+9)+4

Next, distribute the a.

$a(x^2$-6x+9)+4 = $ax^2$ -6ax+9a+4

Notice that in all of our answer choices, the first term is $-2x^2$. If we let a=-2, then we would have $-2x^2$ in our equation. Let's see what happens when we substitute -2 for a.

$f(x)=ax^2$$-6ax+9a+4=(-2)x^2$-6(-2)x+9(-2)+4

$=-2x^2$+12x-18+4