Chapter 2, Problem

Sketch the graph of the quadratic functions in Exercises 1 and 2, indicating the coordinates of the vertex, the y-intercept, and the x-intercepts (if any).

$f\left(x\right)=x^2+2x-3$

To determine

To graph: The function $f\left(x\right)=x^2+2x-3$ and indicate the coordinates of the vertex, the y-intercept, and x-intercept.

Explanation of Solution

Given Information:

The provided function is $f\left(x\right)=x^2+2x-3$.

Graph:

Consider the function,

$f\left(x\right)=x^2+2x-3$

Compare the equation $f\left(x\right)=x^2+2x-3$ with the standard function $f\left(x\right)=a{x}^2+bx+c$ and find the value of $a,b$ and $c$.

The values are $a=1,b=2$ and $c=-3$.

Here $a>0$, therefore the parabola will be upward facing,

To graph a quadratic function, four things should be calculated first.

(i) Vertex

(ii) x-intercept

(iii)y-intercept

(iv) Symmetry

Vertex: The formula of x- coordinate of a vertex is,

$x=\frac{-b}{2a}$

Substitute $a=1$ and $b=2$ in the equation $x=\frac{-b}{2a}$.

$\begin{array}{c}x=\frac{-2}{2\cdot 1}\\ =-\frac{2}{2}\\ =-1\end{array}$

To find the y-coordinate of vertex, substitute $x=-1$ in the function $f\left(x\right)=x^2+2x-3$.

$\begin{array}{c}f\left(x\right)={\left(-1\right)}^2+3\left(-1\right)+2\\ =1-3+2\\ =0\end{array}$

Thus, coordinates of the vertex are $\left(-1,0\right)$.

To calculate the x-intercept of the function, substitute $f\left(x\right)=0$ in the equation $f\left(x\right)=x^2+2x-3$ and solve.

$\begin{array}{c}{x}^2+2x-3=0\\ x^2+3x-x-3=0\\ x\left(x+3\right)-1\left(x+3\right)=0\\ \left(x+3\right)\left(x-1\right)=0\end{array}$

Therefore, $x+3=0$ or $x-1=0$.

Now solve for $x+3=0$.

$\begin{array}{c}x+3=0\\ x=-3\end{array}$

Now solve for $x-1=0$.

$\begin{array}{c}x-1=0\\ x=1\end{array}$

Thus, x-intercept are $-3,1$.

To calculate the y-intercept, substitute $x=0$ in the function $f\left(x\right)=x^2+2x-3$.

$\begin{array}{c}f\left(x\right)={\left(0\right)}^2+2\left(0\right)-3\\ =-3\end{array}$

Thus, the y-intercept is $-3$.

Symmetry:

The formula of the symmetry line is,

$x=-\frac{b}{2a}$

Substitute $a=1$ and $b=2$ in the equation $x=\frac{-b}{2a}$.

$\begin{array}{c}x=\frac{-2}{2\cdot 1}\\ =-\frac{2}{2}\\ =-1\end{array}$

So, the line of symmetry is $x=-1$.

Now plot the graph.

Interpretation:

The graph of $f\left(x\right)=x^2+2x-3$ represents a parabola which is upward facing. Also, the vertex is $\left(-1,-4\right)$, the y-intercept $-3$, and x-intercept are $-3$ and 1.