How to use the quadratic function - Algebra 1

Card 0 of 9

Question

Solve the equation:

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

Answer

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 and to obtain the answers and .

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Question

Solve for :

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

Answer

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:

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Question

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

Answer

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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Question

Which of the following is the correct solution when is solved using the quadratic equation?

Answer

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

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

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

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

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Question

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

Answer

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

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

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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Question

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 ?

Answer

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:

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Question

Solve for x.

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

Answer

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:

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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Question

Solve for :

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

Answer

To find , we must factor the quadratic function:

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

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

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Question

Solve for :

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

Answer

To find , we want to factor the quadratic function:

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

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

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