How to use the quadratic function - PSAT Math

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Question

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

Answer

x2 + 2x - 1 = 7

x2 + 2x - 8 = 0

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

x = -4, x = 2

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Question

Which of the following quadratic equations has a vertex located at $\dpi{100} (3,4)$ ?

Answer

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

$f(x)=a(x-h)^2 +k$ , where the point $\dpi{100} (h,k)$ is the vertex, and $\dpi{100} a$ is a constant.

We are told that the vertex must occur at $\dpi{100} (3,4)$ , so let's plug this information into the vertex form of the equation. $\dpi{100} h$ will be 3, and $\dpi{100} k$ 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 $\dpi{100} a$ .

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

Notice that in all of our answer choices, the first term is $-2x^2$ . If we let $\dpi{100} a=-2$ , then we would have $-2x^2$ in our equation. Let's see what happens when we substitute $\dpi{100} -2$ for $\dpi{100} a$ .

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

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

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Question

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

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

Answer

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,

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

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

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

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

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Question

Define function as follows:

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

Given that and , evaluate .

Answer

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

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

We solve for in the equation

$a^{2} - 10a+ 7 = -18$

$a^{2} - 10a+ 25= 0$

$(a-5)^{2} = 0$

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

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Question

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)

Answer

Set the height function equal to 0:

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

$-16t^{2}+ 90 t + 600 = 0$

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.

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Question

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)

Answer

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.

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Question

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)

Answer

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.

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Question

Define function as follows:

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

Given that and , evaluate .

Answer

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

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

Solve for in the equation

$a^{2} - 6a + 5 = 32$

$a^{2} - 6a -27 = 0$

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

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Question

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)

Answer

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

.

$-16t^{2}+ 90 t + 600 = 600$

$-16t^{2}+ 90 t = 0$

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

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.

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