How do you find the vertex and the intercepts for y=-x^2+4x+12?

Jul 8, 2016

$y$-intercept = $\left(0 , 12\right)$
$x$-intercepts = $\left(6 , 0\right)$ & $\left(- 2 , 0\right)$
Vertex = $\left(2 , 16\right)$

Explanation:

The first stage is to factorise the equation, since ${x}^{2}$ is negative I will multiply the equation by $- 1$ to make it easier to factorise.

Factorise:
$y = {x}^{2} - 4 x - 12$
$\left(x - 6\right) \left(x + 2\right)$

To obtain the $x$-intercept we need to make $y = 0$,

$0 = \left(x - 6\right) \left(x + 2\right)$

We can now solve for the two values

${x}_{1} - 6 = 0$
${x}_{1} = 6$

${x}_{2} + 2 = 0$
${x}_{2} = - 2$

To obtain the axis of symmetry for the parabola we add the two $x$ values together then divide by $2$. This will give the $x$ value for the vertex.

${x}_{v} = \frac{{x}_{1} + {x}_{2}}{2} = \frac{6 - 2}{2} = 2$

Now to get the $y$ value for the vertex substitute $x = 2$ into the original equation and solve:

${y}_{v} = - {x}^{2} + 4 x + 12$
${y}_{v} = {\left(2\right)}^{2} - 4 \left(2\right) - 12$
${y}_{v} = 16$

Therefore the vertex is $\left(2 , 16\right)$

The final intercept we need is the $y$-intercept, this can be calculated by substituting $x = 0$ into the original equation:

$y = - {x}^{2} + 4 x + 12$
$y = - {\left(0\right)}^{2} + 4 \left(0\right) + 12$
$y = 12$

$y$-intercept = $\left(0 , 12\right)$