Q 74. - Precalculus Enhanced with Graphing Utilities

Question

The graph of the function $f\left(x\right)=a{x}^2+bx+c$ has vertex at $\left(1,4\right)$ and passes through the point $\left(-1,-8\right)$. Find  $a, b$ and $c$.

Step-by-Step Solution

Answer

$a=-3, b=6, c=1$

Step 1. Given information.

Given the parabola $f\left(x\right)=a{x}^2+bx+c$.

Given the vertex $\left(1,4\right)$ and a point $\left(-1,-8\right)$. 

Step 2. Determine the value of a .

The vertex of a parabola is $\left(h,k\right)=\left(1,4\right)$. So, $h=1$ and $k=4$.

The vertex form of the parabola is:

$f\left(x\right)=a{\left(x-h\right)}^2+k$

We know that one additional point $\left(-1,-8\right)$.

So, $x=-1$ and $\begin{array}{rcl}f\left(x\right)& =& f\left(-1\right)\\ & =& -8\end{array}$

The required value of $a$ is:

$\begin{array}{rcl}f\left(x\right)& =& a{\left(x-h\right)}^2+k\\ f\left(-1\right)& =& a{\left(-1-1\right)}^2+4\cdots \cdots \cdots \left(x=-1,h=1,k=4,f\left(-1\right)=-8\right)\\ -8& =& a{\left(-2\right)}^2+4\\ -8& =& 4a+4\\ 4a& =& -12\\ a& =& \frac{-12}{4}\\ a& =& -3\end{array}$

Step 3. Determine the value of b and c .

$\begin{array}{rcl}f\left(x\right)& =& a{\left(x-h\right)}^2+k\\ f\left(x\right)& =& -3{\left(x-1\right)}^2+4\cdots \cdots \left(h=1,k=4,a=-3\right)\\ f\left(x\right)& =& -3\left(x^2-2x+1\right)+4\\ f\left(x\right)& =& -3x^2+6x-3+4\\ f\left(x\right)& =& -3x^2+6x+1\end{array}$

Compare the function $f\left(x\right)=-3x^2+6x+1$ with $f\left(x\right)=a{x}^2+bx+c$.

Hence, $a=-3, b=6$ and $c=1$.