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.

Short answer

The value of $a$is$\pm 5.$

Step-by-step solution

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,

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