Finding the roots of a quadratic in vertex form

By Tutor GuyNo Comments


If you have taken algebra 2, you know you can write quadratic functions in three forms:

  • Standard form: y = ax^2 + bx +c
  • Vertex form: y = a(x-h)^2 + k
  • Factored form: y = a(x-r_1)(x-r_2)

No matter which of the forms you have, you are often asked to find the roots (x-intercepts). The factored form is already done for you: the roots are the values of r_1 and r_2 . If your equation is in standard form, you use the quadratic formula. But what do you do if you have a quadratic in vertex form? Most students are taught to expand the equation into standard form (and then use the quadratic formula). But there’s a quick shortcut that is pretty easy to use. In fact, you can write the roots by inspection. If 0 = a(x-h)^2 + k , then the roots are

x = h \pm \sqrt{\dfrac{-k}{a}}

There are three cases:

  • k=0 : There is one (double) root;
  • a \text{ and } k have different signs: the two roots are real;
  • a \text{ and } k have the same sign: the two roots are complex.

In the complex case, you can express the roots as

x = h \pm \sqrt{\dfrac{k}{a}} \; i

Here are two quick examples:

Example 1: 0 = 2(x-3)^2 + 6

Using the formula above, the roots can be written immediately as

x = 3 \pm \sqrt{\dfrac{6}{2}} \; i =3 \pm \sqrt{3} \; i

Example 2: 0 = 2(x-3)^2 - 8

The roots are:

x = 3 \pm \sqrt{\dfrac{8}{2}} = 3 \pm 2 = 1 \text{ and } 5

Algebra 2, Precalc/Trig
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