Question

Find the correct statement pertaining to the functions $f\left(x\right)={(x-3)}^2+2$ and $g\left(x\right)=\frac{1}{2}x+1$ graphed above

• A. $f\left(x\right)=g\left(x\right)$ for exactly $2$ values of x
• B. $f\left(x\right)=g\left(x\right)$ for exactly $1$ values of x
• C. $f\left(x\right)$ is the inverse of $g\left(x\right)$
• D. $f\left(x\right)>g\left(x\right)$ for all $x$

Answer 1

Answer: (A)

$f\left(x\right)=g\left(x\right)$ for exactly $2$ values of x

If $f\left(x\right)=(x-3{)}^2+2$ and $g\left(x\right)=\frac{1}{2}x+1$ then $f\left(x\right)=g\left(x\right)$ is as follows:

$(x-3{)}^2+2=\frac{1}{2}x+1$

$x^2+9-6x+2=\frac{x+2}{2}$

$x^2-6x+11=\frac{x+2}{2}$

$2(x^2-6x+11)=x+2$

$2x^2-12x+22=x+2$

$2x^2-13x+20=0$

Now finding the roots of the quadratic equation $2x^2-13x+20=0$:

$2x^2-13x+20=0$

$2x^2-8x-5x+20=0$

$2x(x-4)-5(x-4)=0$

$2x-5=0,x-4=0$

$x=\frac{5}{2},x=4$

Hence $f\left(x\right)=g\left(x\right)$ for exactly $2$ values of $x$.