If $f(x)=-2x^{2}+7$ and $g(x)=|x-1|$ , find all values of $x$ , to the nearest tenth, for which $f(x)=g(x)$ .

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Algebra 2

Transformations of absolute value functions

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Q. If

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

and

g(x)=|x-1|

, find all values of

x

, to the nearest tenth, for which

f(x)=g(x)

Set Equations Equal: To find the values of $x$ for which $f(x) = g(x)$ , we need to set the two functions equal to each other and solve for $x$ . $f(x) = g(x)$ $-2x^2 + 7 = |x - 1|$
Consider Cases: We need to consider two cases for the absolute value function: when $x - 1$ is non-negative and when $x - 1$ is negative. $\newline$ Case $1$ : $x - 1 \geq 0$ , which implies $x \geq 1$ $\newline$ $-2x^2 + 7 = x - 1$
Solve Case $1$ Quadratic: Now we solve the quadratic equation for Case $1$ by moving all terms to one side of the equation. $\newline$ $-2x^2 - x + 8 = 0$
Calculate Roots Case $1$ : We can use the quadratic formula to find the roots of the equation. The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = -2$ , $b = -1$ , and $c = 8$ . $x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(-2)(8)}}{2(-2)}$
Find $x_1$ and $x_2$ Case $1$ : Calculate the discriminant $(b^2 - 4ac)$ and then the roots. $\newline$ Discriminant = $(-1)^2 - 4(-2)(8) = 1 + 64 = 65$ $\newline$ $x = \frac{1 \pm \sqrt{65}}{-4}$
Check Validity Case $1$ : Now we find the two roots for Case $1$ . $\newline$ $x_1 = \frac{1 + \sqrt{65}}{-4}$ $\newline$ $x_2 = \frac{1 - \sqrt{65}}{-4}$
Solve Case $2$ Quadratic: Calculate the numerical values of $x_1$ and $x_2$ to the nearest tenth. $\newline$ $x_1 \approx (1 + 8.0623) / (-4) \approx 9.0623 / (-4) \approx -2.3$ $\newline$ $x_2 \approx (1 - 8.0623) / (-4) \approx -7.0623 / (-4) \approx 1.8$
Calculate Roots Case $2$ : We must check if these solutions are valid for Case $1$ , which requires $x \geq 1$ . $x_1 = -2.3$ is not valid because it is less than $1$ . $x_2 = 1.8$ is valid because it is greater than or equal to $1$ .
Find $x_1$ and $x_2$ Case $2$ : Now we consider Case $2$ : $x - 1 < 0$ , which implies $x < 1$ $-2x^2 + 7 = -(x - 1)$
Check Validity Case $2$ : Solve the quadratic equation for Case $2$ by moving all terms to one side of the equation. $\newline$ $-2x^2 + 7 = -x + 1$ $\newline$ $-2x^2 + x + 6 = 0$
Combine Valid Solutions: We can use the quadratic formula to find the roots of the equation for Case $2$ . The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = -2$ , $b = 1$ , and $c = 6$ . $\newline$ $x = \frac{-(1) \pm \sqrt{(1)^2 - 4(-2)(6)}}{2(-2)}$
Combine Valid Solutions: We can use the quadratic formula to find the roots of the equation for Case $2$ . The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = -2$ , $b = 1$ , and $c = 6$ .
$x = \frac{-(1) \pm \sqrt{(1)^2 - 4(-2)(6)}}{2(-2)}$ Calculate the discriminant ( $b^2 - 4ac$ ) and then the roots for Case $2$ .
Discriminant = $(1)^2 - 4(-2)(6) = 1 + 48 = 49$
$x = \frac{-1 \pm \sqrt{49}}{-4}$
Combine Valid Solutions: We can use the quadratic formula to find the roots of the equation for Case $2$ . The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = -2$ , $b = 1$ , and $c = 6$ .
$x = \frac{-(1) \pm \sqrt{(1)^2 - 4(-2)(6)}}{2(-2)}$ Calculate the discriminant ( $b^2 - 4ac$ ) and then the roots for Case $2$ .
Discriminant = $(1)^2 - 4(-2)(6) = 1 + 48 = 49$
$x = \frac{-1 \pm \sqrt{49}}{-4}$ Now we find the two roots for Case $2$ .
$x_1 = \frac{-1 + \sqrt{49}}{-4}$
$x_2 = \frac{-1 - \sqrt{49}}{-4}$
Combine Valid Solutions: We can use the quadratic formula to find the roots of the equation for Case $2$ . The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = -2$ , $b = 1$ , and $c = 6$ .
$x = \frac{-(1) \pm \sqrt{(1)^2 - 4(-2)(6)}}{2(-2)}$ Calculate the discriminant ( $b^2 - 4ac$ ) and then the roots for Case $2$ .
Discriminant = $(1)^2 - 4(-2)(6) = 1 + 48 = 49$
$x = \frac{-1 \pm \sqrt{49}}{-4}$ Now we find the two roots for Case $2$ .
$x_1 = \frac{-1 + \sqrt{49}}{-4}$
$x_2 = \frac{-1 - \sqrt{49}}{-4}$ Calculate the numerical values of $a = -2$ $0$ and $a = -2$ $1$ to the nearest tenth for Case $2$ .
$a = -2$ $2$
$a = -2$ $3$
Combine Valid Solutions: We can use the quadratic formula to find the roots of the equation for Case $2$ . The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = -2$ , $b = 1$ , and $c = 6$ .
$x = \frac{-(1) \pm \sqrt{(1)^2 - 4(-2)(6)}}{2(-2)}$ Calculate the discriminant ( $b^2 - 4ac$ ) and then the roots for Case $2$ .
Discriminant = $(1)^2 - 4(-2)(6) = 1 + 48 = 49$
$x = \frac{-1 \pm \sqrt{49}}{-4}$ Now we find the two roots for Case $2$ .
$x_1 = \frac{-1 + \sqrt{49}}{-4}$
$x_2 = \frac{-1 - \sqrt{49}}{-4}$ Calculate the numerical values of $a = -2$ $0$ and $a = -2$ $1$ to the nearest tenth for Case $2$ .
$a = -2$ $2$
$a = -2$ $3$ We must check if these solutions are valid for Case $2$ , which requires $a = -2$ $4$ .
$a = -2$ $5$ is valid because it is less than $a = -2$ $6$ .
$a = -2$ $7$ is not valid because it is greater than or equal to $a = -2$ $6$ .
Combine Valid Solutions: We can use the quadratic formula to find the roots of the equation for Case $2$ . The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = -2$ , $b = 1$ , and $c = 6$ .
$x = \frac{-(1) \pm \sqrt{(1)^2 - 4(-2)(6)}}{2(-2)}$ Calculate the discriminant ( $b^2 - 4ac$ ) and then the roots for Case $2$ .
Discriminant = $(1)^2 - 4(-2)(6) = 1 + 48 = 49$
$x = \frac{-1 \pm \sqrt{49}}{-4}$ Now we find the two roots for Case $2$ .
$x_1 = \frac{-1 + \sqrt{49}}{-4}$
$x_2 = \frac{-1 - \sqrt{49}}{-4}$ Calculate the numerical values of $a = -2$ $0$ and $a = -2$ $1$ to the nearest tenth for Case $2$ .
$a = -2$ $2$
$a = -2$ $3$ We must check if these solutions are valid for Case $2$ , which requires $a = -2$ $4$ .
$a = -2$ $5$ is valid because it is less than $a = -2$ $6$ .
$a = -2$ $7$ is not valid because it is greater than or equal to $a = -2$ $6$ .Combine the valid solutions from both cases.
From Case $1$ , we have $a = -2$ $9$ .
From Case $2$ , we have $b = 1$ $0$ .
These are the values of $b = 1$ $1$ for which $b = 1$ $2$ .