Chapter 3.3, Problem 51AYU

In Problems 49-54, determine the quadratic function whose graph is given.

To determine

To calculate: The quadratic function using the given graphs.

Answer to Problem 51AYU

The equation of the given graph is $f(x)=2x^2-4x-1$.

Explanation of Solution

Given:

The given graph is

Formula Used:

The general form of a quadratic equation is $f(x)=a{(x-h)}^2+k,a\ne 0$.

This general equation can also be written as $f(x)=a{x}^2+bx+c,a\ne 0$, where

$\begin{array}{l}h=-\frac{b}{2a}\\ \text{and}\\ k=f\left(-\frac{b}{2a}\right)\end{array}$

If $a>0$, the graph opens upwards.

If $a<0$, the graph opens downwards.

The vertex of the above function is $(h,k)=\left(\frac{-b}{2a},f\left(\frac{-b}{2a}\right)\right)$.

The axis of symmetry will be $x=\frac{-b}{2a}$.

We can find the $y\text{-intercept}$ by equating the equation at $x=0$.

We can find the $x\text{-intercept}$ by equation the equation at $y=0$.

1. If $b^2-4ac=0$ then the vertex is the $x\text{-intercept}$.

2. If $b^2-4ac\text{ < }0$ then the graph has no $x\text{-intercept}$.

3. If $b^2-4ac\text{ > }0$ then the vertex is the $x\text{-intercept}$.

Calculation:

Let us consider the quadratic equation of the given graph to be

$f(x)=a{x}^2+bx+c,a\ne 0$-----(1)

From the given graph, we can see that the $y\text{-intercept}$ of the given function is $(0,-1)$.

We know that the $y\text{-intercept}$ is the value of the function at $x=0$, thus, we have

$f(0)=\text{ -1}$-----(2)

At $x=0$, (1) becomes

$f(0)=c$-----(3)

Thus, using (2) and (3), we get the value of $c=-1$.

Thus, equation (1) becomes

$f(x)=a{x}^2+bx-1$-----(4)

Now, we need to find $a$ and $b$.

We know that a point in a graph is written as $\left(x,f(x)\right)$.

The given graph has 2 points $(-1,5)$ and $(3,5)$.

Consider the point $(-1,5)$.

Here, we have $x=-1$ and $f(x)=5$.

Therefore, substituting it in (4), we get

$\begin{array}{l}5=a{\left(-1\right)}^2+b(-1)-1\\ \Rightarrow a-b=6\end{array}$-----(5)

Now, consider the point $(3,5)$.

Here, we have $x=\text{ 3}$ and $f(x)=5$.

Therefore, substituting it in (4), we get

$\begin{array}{l}5=a{\left(3\right)}^2+b(3)-1\\ \Rightarrow 9a+3b=6\\ \Rightarrow 3a+b=2\end{array}$-----(6)

In order to find the values of $a$ and $b$, we have to solve the equations (5) and (6).

Therefore, we have

$a-b=6$-----(5) And

$3a+b=2$-----(6).

Now, on adding (5) and (6), we get

$\begin{array}{l}4a=8\\ \Rightarrow a=2\end{array}$

On substituting the above value in (5), we get

$\begin{array}{l}2-b=6\\ \Rightarrow 2-6=b\\ \Rightarrow b=-4\end{array}$

Therefore, the equation of the given graph is

$f(x)=2x^2-4x-1$