Q 73. - Precalculus Enhanced with Graphing Utilities

Question

The graph of the function $f\left(x\right)=a{x}^2+bx+c$ has vertex at $\left(0,2\right)$ and passes through the point $\left(1,8\right)$. Find $a,b$ and $c$.

Step-by-Step Solution

Answer

The solutions are $a=6,b=0$, and $c=2$.

Step 1. Given information.

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

Vertex at $\left(0,2\right)$ and passes through the point $\left(1,8\right)$.

Step 2. Determine the value of b and c .

The coordinates of the vertex of a quadratic function of the form $f\left(x\right)=a{x}^2+bx+c$ are $\left(\frac{-b}{2a},f\left(\frac{-b}{2a}\right)\right)$.

So, $\left(\frac{-b}{2a},f\left(\frac{-b}{2a}\right)\right)=\left(0,2\right)\cdots \cdots \left(Given vertex \left(0,2\right)\right)$.

$\begin{array}{rcl}\frac{-b}{2a}& =& 0\\ -b& =& 0\\ b& =& 0\end{array}$       and     $\begin{array}{rcl}f\left(\frac{-b}{2a}\right)& =& 2\\ f\left(0\right)& =& 2\cdots \cdots \left(\frac{-b}{2a}=0\right)\\ a{\left(0\right)}^2+b\left(0\right)+c& =& 2\cdots \cdots \left(substitute x=0 in f\left(x\right)=a{x}^2+bx+c\right)\\ c& =& 2\end{array}$

Step 3. Determine the value of a .

From the point $\left(1,8\right)$, we find that $f\left(1\right)=8$.

Substitute $1$ for $x$, $0$ for $b$, $2$ for $c$, and $8$ for $f\left(1\right)$ in $f\left(x\right)=a{x}^2+bx+c$.

$\begin{array}{rcl}f\left(x\right)& =& a{x}^2+bx+c\\ f\left(1\right)& =& a{\left(1\right)}^2+0\left(1\right)+2\\ 8& =& a+2\\ a+2& =& 8\\ a& =& 8-2\\ a& =& 6\end{array}$

Step 4. Simplified answer.

Hence, $a=6,b=0$, and $c=2$.