Example: Finding the y- and x-Intercepts of a Parabola

Find the y- and x-intercepts of the quadratic [latex]f\left(x\right)=3{x}^{2}+5x - 2[/latex].

Answer: We find the y-intercept by evaluating [latex]f\left(0\right)[/latex].

[latex]\begin{array}{c}f\left(0\right)=3{\left(0\right)}^{2}+5\left(0\right)-2\hfill \\ \text{ }=-2\hfill \end{array}[/latex]

So the y-intercept is at [latex]\left(0,-2\right)[/latex]. For the x-intercepts, or roots, we find all solutions of [latex]f\left(x\right)=0[/latex].

[latex]0=3{x}^{2}+5x - 2[/latex]

In this case, the quadratic can be factored easily, providing the simplest method for solution.

[latex]0=\left(3x - 1\right)\left(x+2\right)[/latex] [latex]\begin{array}{c}0=3x - 1\hfill & \hfill & \hfill & \hfill & 0=x+2\hfill \\ x=\frac{1}{3}\hfill & \hfill & \text{or}\hfill & \hfill & x=-2\hfill \end{array}[/latex]

So the roots are at [latex]\left(\frac{1}{3},0\right)[/latex] and [latex]\left(-2,0\right)[/latex].

Analysis of the Solution

By graphing the function, we can confirm that the graph crosses the y-axis at [latex]\left(0,-2\right)[/latex]. We can also confirm that the graph crosses the x-axis at [latex]\left(\frac{1}{3},0\right)[/latex] and [latex]\left(-2,0\right)[/latex].

Example: Finding the Roots of a Parabola

Find the x-intercepts of the quadratic function [latex]f\left(x\right)=2{x}^{2}+4x - 4[/latex].

Answer: We begin by solving for when the output will be zero.

[latex]0=2{x}^{2}+4x - 4[/latex]

Because the quadratic is not easily factorable in this case, we solve for the intercepts by first rewriting the quadratic in standard form.

[latex]f\left(x\right)=a{\left(x-h\right)}^{2}+k[/latex]

We know that a = 2. Then we solve for h and k.

[latex]\begin{array}{c}h=-\frac{b}{2a}\hfill & \hfill & \hfill & k=f\left(-1\right)\hfill \\ \text{ }=-\frac{4}{2\left(2\right)}\hfill & \hfill & \hfill & \text{ }=2{\left(-1\right)}^{2}+4\left(-1\right)-4\hfill \\ \text{ }=-1\hfill & \hfill & \hfill & \text{ }=-6\hfill \end{array}[/latex]

So now we can rewrite in standard form.

[latex]f\left(x\right)=2{\left(x+1\right)}^{2}-6[/latex]

We can now solve for when the output will be zero.

[latex]\begin{array}{c}0=2{\left(x+1\right)}^{2}-6\hfill \\ 6=2{\left(x+1\right)}^{2}\hfill \\ 3={\left(x+1\right)}^{2}\hfill \\ x+1=\pm \sqrt{3}\hfill \\ x=-1\pm \sqrt{3}\hfill \end{array}[/latex]

The graph has x-intercepts at [latex]\left(-1-\sqrt{3},0\right)[/latex] and [latex]\left(-1+\sqrt{3},0\right)[/latex].

Analysis of the Solution

We can check our work by graphing the given function on a graphing utility and observing the roots.