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Exercises Polynomials and factoring

Example 8 from Algebra and Trigonometry, 5.1 Quadratic Functions

Finding the x-Intercepts of a Parabola

Find the $x -$ intercepts of the quadratic function $f (x) = 2 x^{2} + 4 x - 4.$

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

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

$0 = 2 x^{2} + 4 x - 4$

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

$f (x) = a {(x - h)}^{2} + k$

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

$\begin{matrix} h & = & - \frac{b}{2 a} & k & = & f (−1) \\ = & - \frac{4}{2 (2)} & = & 2 {(−1)}^{2} + 4 (−1) - 4 \\ = & −1 & = & -6 \end{matrix}$
So now we can rewrite in standard form.

$f (x) = 2 {(x + 1)}^{2} - 6$

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

$\begin{array}{l} 0 = 2 {(x + 1)}^{2} - 6 \\ 6 = 2 {(x + 1)}^{2} \\ 3 = {(x + 1)}^{2} \\ x + 1 = \pm \sqrt{3} \\ x = - 1 \pm \sqrt{3} \end{array}$

The graph has x-intercepts at $(−1 - \sqrt{3}, 0)$ and $(−1 + \sqrt{3}, 0) .$
We can check our work by graphing the given function on a graphing utility and observing the $x -$ intercepts. See Figure 15.

Figure 15

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

In a Try It, we found the standard and general form for the function $g (x) = 13 + x^{2} - 6 x .$ Now find the y- and x-intercepts (if any).

How did it go?

PreviousFinding the y- and x-Intercepts of a Parabola Next Applying the Vertex and x-Intercepts of a Parabola

From Algebra and Trigonometry (1st edition) by OpenStax (Rice University), section 5.1, licensed under CC BY 4.0. Original at openstax.org. We split the solution into steps; the words, numbers and figures are the book's. OpenStax doesn't endorse this site.