Q. 61 - Precalculus Enhanced with Graphing Utilities

Question

If $f\left(x\right)=2x^2+5$and $g\left(x\right)=3x+a$ , find $a$ so that the graph of $f\circ g$ crosses the y-axis at 23. 

Step-by-Step Solution

Answer

The value of $a$ is $\pm 3.$

Step 1. Given Information

Given that $f\left(x\right)=2x^2+5$ and  $g\left(x\right)=3x+a,$ value of $f\circ g\left(x\right)=23$ at $x=0.$

Step 2. Solution

We know that $f\circ g\left(x\right)=f\left(g\right(x\left)\right).$

Here, $f\left(x\right)=2x^2+5$ and $g\left(x\right)=3x+a.$

So, $$\begin{gathered} f\circ g\left(x\right)=f\left(g\right(x\left)\right) \\ f\circ g\left(x\right)=2(g{\left(x\right))}^2+5f\circ g(x)=2{(3x+a)}^2+5f\circ g(x)=2\{9x^2+6xa+a^2\}+5 \\ f\circ g(x)=18x^2+12xa+2a^2+5 \end{gathered}$$

So, at $x=0$, $f\circ g\left(x\right)=23.$

Now,

 $$\begin{gathered} at x=0, \\ f\circ g\left(0\right)=18{\left(0\right)}^2+12\left(0\right)\left(a\right)+2a^2+5 \\ 23=0+0+2a^2+5 \\ 2a^2=18 \\ {a}^2=9 \\ a=\pm 3. \end{gathered}$$