What is the the vertex of y = (x-1)^2+x-12?

Jul 6, 2017

Vertex is $\left(\frac{1}{2} , - 11 \frac{1}{4}\right)$

Explanation:

Vertex form of equation is $y = a {\left(x - h\right)}^{2} + k$, where $\left(h , k\right)$ is the vertex.

Now $y = {\left(x - 1\right)}^{2} + x - 12$

= ${x}^{2} - 2 x + 1 + x - 12$

= ${x}^{2} - x - 11$

= ${x}^{2} - 2 \times \frac{1}{2} \times x + {\left(\frac{1}{2}\right)}^{2} - \frac{1}{4} - 11$

= ${\left(x - \frac{1}{2}\right)}^{2} - \frac{45}{4}$

Hence $y = {\left(x - 1\right)}^{2} + x - 12 \Leftrightarrow y = {\left(x - \frac{1}{2}\right)}^{2} - \frac{45}{4}$

and vertex is $\left(\frac{1}{2} , - 11 \frac{1}{4}\right)$

graph{(x-1)^2+x-12 [-19.83, 20.17, -14.16, 5.84]}