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

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Answer 1 · 2016-07-08T03:39:46

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

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:
$y = x^2 - 4x - 12$
$(x - 6)(x + 2)$

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

$0 = (x - 6)(x+2)$

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.

$x_v = (x_1 + x_2)/ 2= (6 - 2) / 2 = 2$

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

$y_v = -x^2 + 4x +12$
$y_v = (2)^2 - 4(2) -12$
$y_v = 16$

Therefore the vertex is $(2,16)$

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

$y = -x^2 + 4x +12$
$y = -(0)^2 + 4(0) +12$
$y = 12$

$y$ -intercept = $(0,12)$

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