Q. 62 - Precalculus Enhanced with Graphing Utilities

Question

If $f\left(x\right)=3x^2-7$ and $g\left(x\right)=2x+a$, find $a$ so that the graph of $f\circ g$ crosses the y-axis at 68.

Step-by-Step Solution

Answer

The value of $a$ is $\pm 5.$

Step 1. Given Information

Given that $f\left(x\right)=3x^2-7$ and $g\left(x\right)=2x+a,$ value of $f\circ g\left(x\right)$ at $x=0$ is 68.

Step 2. Solution

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

Here, $f\left(x\right)=3x^2-7, g\left(x\right)=2x+a$

$$\begin{gathered} f\circ g\left(x\right)=f\left(g\right(x\left)\right) \\ f\circ g\left(x\right)=3(g{\left(x\right))}^2-7f\circ g(x)=3{\{2x+a\}}^2-7f\circ g(x)=3\{4x^2+4xa+a^2\}-7 \\ f\circ g(x)=12x^2+12xa+3a^2-7 \end{gathered}$$

So, at $x=0$, $f\circ g\left(0\right)=68.$

Now,

$$\begin{gathered} f\circ g\left(x\right)=12x^2+12xa+3a^2-7, \\ at x=0 \\ f\circ g\left(0\right)=12{\left(0\right)}^2+12\left(0\right)\left(a\right)+3a^2-7 \\ 68=0+0+3a^2-7 \\ 3a^2=68+7 \\ 3a^2=75 \\ {a}^2=25 \\ a=\pm 5. \end{gathered}$$