# How do you find the vertex and intercepts for 12y = x^2 – 6x + 45?

##### 1 Answer
Mar 30, 2017

Vertex is $\left(3 , 3\right)$, $y$ intercept is $3 \frac{3}{4}$ and there is no $x$-intercept.

#### Explanation:

In an equation $y = a {\left(x - h\right)}^{2} + k$, we have vertex as $\left(h , k\right)$ and intercepts on $y$-axis can be found by putting $x = 0$ and intercepts on $x$-axis can be found by putting $y = 0$. Let us convert the given equation in vertex form.

$12 y = {x}^{2} - 6 x + 45$ can be written as

$y = \frac{1}{12} \left({x}^{2} - 6 x\right) + \frac{45}{12}$

or $y = \frac{1}{12} \left({x}^{2} - 2 \times 3 \times x + {3}^{2} - {3}^{2}\right) + \frac{45}{12}$

or $y = \frac{1}{12} \left({x}^{2} - 2 \times 3 \times x + {3}^{2}\right) - \frac{9}{12} + \frac{45}{12}$

or $y = \frac{1}{12} {\left(x - 3\right)}^{2} + \frac{36}{12}$

or $y = \frac{1}{12} {\left(x - 3\right)}^{2} + 3$

Hence vertex is $\left(3 , 3\right)$

intercept on $y$-axis is $y = \frac{15}{4} = 3 \frac{3}{4}$

Observe that as $a$ and $k$ are positive, $y$ is always greater than $0$ and can never be $0$, hence no $x$-intercept.
graph{12y=x^2-6x+45 [-7.12, 12.88, -0.4, 9.6]}