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Algebra 1 : How to use the quadratic function

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Example Questions

Example Question #2 : Finding Roots

Solve the equation:

$\small 13 = x^2-12x$

Possible Answers:

$\small \left \{-1, 13 \right \}$

$\small \left \{0\right\}$

$\small \left \{ 2, 4\right \}$

$\small \left \{ 1, 2 \right \}$

$\small \left \{-13, 1\right \}$

Correct answer:

$\small \left \{-1, 13 \right \}$

Explanation:

To solve the quadratic equation, $\small 13= x^{2} -12x$ , we set the equation equal to zero and then factor the quadratic, $\small (x-13)(x+1)=0$ . Because these expressions multiply to equal 0, then it must be that at least one of the expressions equals 0. So we set up the corresponding equations x-13=0 and x+1=0 to obtain the answers x=13 and x=-1 .

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Example Question #2 : Quadratic Formula

Solve for :

$x^{2}+12x+35=0$

Possible Answers:

x=3,15

x=-5,-7

x=-5,-9

x=7,9

The solution is undefined.

Correct answer:

x=-5,-7

Explanation:

To factor this equation, first find two numbers that multiply to 35 and sum to 12. These numbers are 5 and 7. Split up 12x using these two coefficients:

x^2 + 7x+5x+35=0

x(x+7)+5(x+7)=0

(x+5)(x+7)=0

x=-5,-7

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Example Question #1 : How To Use The Quadratic Function

Given $f(x) = -2x^{2} - 3x +5$ , find f(a+h) - f(a) .

Possible Answers:

$2a^{2}+3a-5$

f(a+h) -f(a)

$4ah+2h^{2} -3h$

$-2a^{2}-4ah-2h^{2} -3a -3h +5$

$-2h^{2} - 4ah -3h$

Correct answer:

$-2h^{2} - 4ah -3h$

Explanation:

Plug in a for x:

$f(a)=-(-2a^{2}-3a+5) = 2a^2+3a-5$

Next plug in (a + h) for x:

$f(a+h) = -2(a+h)^2-3(a+h)+5 = -2a^{2} - 4ah - 2h^{2} -3a - 3h +5$

Therefore f(a+h) - f(a) = $-2h^{2} - 4ah - 3h$ .

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Example Question #2 : How To Use The Quadratic Function

Which of the following is the correct solution when x^2-10x-11=0 is solved using the quadratic equation?

Possible Answers:

$\frac{-10\pm 12}{2}$

$10\pm 12$

$\frac{10\pm \sqrt{56}}{2}$

$\frac{10\pm 12}{2}$

Correct answer:

$\frac{10\pm 12}{2}$

Explanation:

$x=\frac{-(-10)\pm \sqrt{(-10)^2-4(1)(-11)}}{2(1)}$

$=\frac{10\pm \sqrt{100+44}}{2}$

$=\frac{10\pm \sqrt{144}}{2}$

$=\frac{10\pm 12}{2}$

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Example Question #1 : Parabolic Functions

Give the minimum value of the function $f(x) = 2 x^{2} - 5x + 9$ .

Possible Answers:

$5\frac{7}{8}$

$18\frac{3}{8}$

This function does not have a minimum.

Correct answer:

$5\frac{7}{8}$

Explanation:

This is a quadratic function. The -coordinate of the vertex of the parabola can be determined using the formula $x = - \frac{b}{2a}$ , setting a = 2,b = -5 :

$x = - \frac{b}{2a} = - \frac{-5}{2 \cdot 2} =\frac{5}{4}$

Now evaluate the function at $x = \frac{5}{4}$ :

$f(x) = 2 x^{2} - 5x + 9$

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

$= 2 \left (\frac{25}{16} \right ) - 5 \left ( \frac{5}{4} \right )+ 9$

$= \frac{25}{8} - \frac{25}{4} + 9$

$= \frac{25}{8} - \frac{50}{8} + \frac{72}{8} =\frac{47}{8} =5\frac{7}{8}$

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Example Question #2 : How To Use The Quadratic Function

Quadratic equations may be written in the following format:

$ax^{2}+bx+c$

In the equation , $4x^{2}-4x+2$ what is the value of ?

Possible Answers:

Correct answer:

Explanation:

when using the quadratic formula, your variables are as follows

$ax^{2}+bx+c$

For the given equation below:

$4x^{2}-4x+2$

The values of each coefficient are:

a=4

b=-4

c=2

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Example Question #3 : How To Use The Quadratic Function

Solve for x.

$-15 + 3x^{2}+4x = -5 - x$

Possible Answers:

$\frac{-6\pm \sqrt{25}}{5}$

$\frac{-5\pm \sqrt{120}}{3}$

$\frac{5\pm \sqrt{25}}{6}$

$\frac{-5\pm \sqrt{145}}{6}$

Correct answer:

$\frac{-5\pm \sqrt{145}}{6}$

Explanation:

The quadratic formula is as follows:

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

We will start by finding the values of the coefficients of the given equation, but first we must simplify.

$-15 + 3x^{2}+4x = -5 - x$

Move all the terms to one side and set the equation equal to .

$-15+5 + 3x^{2}+4x+x = 0$

$-10+3x^{2}+5x=0$

Rearrange.

$3x^{2}+5x-10=0$

We can then find the values of the coefficients of the equation:

$3x^{2}+5x-10$

Quadratic equations may be written in the following format:

$ax^{2}+bx+c$

In our case, the values of the coefficients are:

a=3

b=5

c=-10

Substitute the coefficient values into the quadratic equation:

$x=\frac{-5\pm \sqrt{5^{2}-4\left ( 3 \right )\left ( -10 \right )}}{2\left ( 3 \right )}$

$x=\frac{-5\pm \sqrt{25-(-120)}}{6}$

After simplifying we are left with:

$x=\frac{-5\pm \sqrt{145}}{6}$

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Example Question #1 : Functions

Solve for :

$3x^{2}-18x+24=0$

Possible Answers:

$x=2,\,3$

$x=2,\,4$

$x=-1,\,-6$

$x=-6,\,1$

$x=-2,\,4$

Correct answer:

$x=2,\,4$

Explanation:

To find , we must factor the quadratic function:

$3x^{2}-18x+24=0$

$3(x^{2}-6x+8)=0$

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

$x-4=0\, and \, x-2=0$

$x=4\,and\,2$

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Example Question #1 : Understand Functions: Ccss.Math.Content.8.F.A.1

Solve for :

$2x^{2}-10x-28=0$

Possible Answers:

$x=-1\,and\,x=5$

$x=2\,and\,x=-5$

$x=2\,and\,x=-7$

$x=-2\,and\,x=7$

$x=-2\,and\,x=5$

Correct answer:

$x=-2\,and\,x=7$

Explanation:

To find , we want to factor the quadratic function:

$2x^{2}-10x-28=0$

$2(x^{2}-5x-14)=0$

2(x-7)(x+2)=0

$x-7=0\,and\,x+2=0$

$x=7\:and\:-2$

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Algebra 1 : How to use the quadratic function

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

Example Question #2 : Finding Roots

Example Question #2 : Quadratic Formula

Example Question #1 : How To Use The Quadratic Function

Example Question #2 : How To Use The Quadratic Function

Example Question #1 : Parabolic Functions

Example Question #2 : How To Use The Quadratic Function

Example Question #3 : How To Use The Quadratic Function

Example Question #1 : Functions

Example Question #1 : Understand Functions: Ccss.Math.Content.8.F.A.1

All Algebra 1 Resources

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