Question

Determine the quadratic function whose graph is given.

Short answer

The quadratic function is$f\left(x\right)=2x^2-4x-1$.

Step-by-step solution

Step 1. Given information.

The given graph is:

Step 2. Determine the vertex from the given graph.

By analyzing the graph of the function, the vertex of the function is $\left(1,-3\right)$.

If the vertex $\left(h,k\right)$and one additional point is given on the graph of a quadratic function $f\left(x\right)=a{x}^2+bx+c$, then it can be written as $f\left(x\right)=a{\left(x-h\right)}^2+k$.

So, here $\left(h,k\right)=\left(1,-3\right)$.

Thus,$h=1$and$k=-3$.

Step 3. Simplify fx=ax-h2+k by substituting h=1 and k=-3.

$$\begin{gathered} f\left(x\right)=a{\left(x-h\right)}^2+k \\ f\left(x\right)=a{\left(x-1\right)}^2-3\cdots \cdots \left(i\right) \end{gathered}$$

Step 4. By using the additional point 3,5 determine the value of a.

Substitute $x=3$and $f\left(3\right)=5$in the equation (i).

$\begin{array}{rcl}f\left(3\right)& =& a{\left(3-1\right)}^2-3\\ 5& =& a{\left(2\right)}^2-3\\ 5& =& 4a-3\\ 4a-3& =& 5\\ 4a& =& 5+3\\ 4a& =& 8\\ a& =& \frac{8}{4}\\ a& =& 2\end{array}$

Step 5. Now, substitute the value of a=2 in fx=ax-12-3.

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

Step 6. Simplified answer.

Hence, the required quadratic equation is$f\left(x\right)=2x^2-4x-1$.