What is the vertex of y=3x^2+12x - 15?

Jan 5, 2016

Complete the square to convert to vertex form.

Explanation:

y = $3 {x}^{2}$ + 12x - 15
y = 3(${x}^{2}$ + 4x + n - n) - 15
n = ${\left(\frac{b}{2}\right)}^{2}$
n = 4
y = 3(${x}^{2}$ + 4x + 4 - 4) - 15
y = 3(${x}^{2}$ + 4x + 4) - 12 - 15
y = 3(${x}^{2}$ + 4x + 4) - 27
y = 3${\left(x + 2\right)}^{2}$ - 27

In the form y = a${\left(x - p\right)}^{2}$ + q, the vertex can be found at (p, q). So, the vertex is (-2, -27).

Hopefully my explanation helps!