# How can I find the horizontal intercept of the equation y=-4x^2-8x+12 algebraically?

Jun 20, 2018

The horizontal intercepts have coordinates $\left(1 , 0\right)$ and $\left(- 3 , 0\right)$

#### Explanation:

We want to find the intercepts of a certain graph with the $x$ axis. Every point on the $x$ axis has the form $\left(x , 0\right)$, while every point on the graph has the form $\left(x , f \left(x\right)\right)$

This means that a point on the graph is on the $x$ axis if and only if $f \left(x\right) = 0$, so that $\left(x , f \left(x\right)\right) = \left(x , 0\right)$ is a point on the axis.

Since $y = f \left(x\right)$, we must ask that $y = 0$ and solve for $x$: the equation becomes

$0 = - 4 {x}^{2} - 8 x + 12$

or, if you prefer,

$4 {x}^{2} + 8 x - 12 = 0$

note that the whole equation can be simplified dividing by $4$:

${x}^{2} + 2 x - 3 = 0$

To solve this equation, you can either use the quadratic formula, or look for two number that give $- 2$ when summed and $- 3$ when multiplied.

The second request is satisfied only by $1$ and $- 3$ or $- 1$ and $- 3$. Of these two couples, only the first sum to $- 2$, so the solutions are $1$ and $- 3$

Jun 20, 2018

x-intercepts:

$x = - 3$ and $x = 1$

y-intercept:

$y = 12$

#### Explanation:

I assume by "horizontal" you mean the x-intercept?

To find the x-intercept set $y = 0$ and solve for $x$:

$y = - 4 {x}^{2} - 8 x + 12$

$0 = - 4 {x}^{2} - 8 x + 12$

Factor:

$0 = - 4 \left(x + 3\right) \left(x - 1\right)$

$x = - 3$ and $x = 1$

To find the y-intercept set $x = 0$ and solve for $y$:

$y = - 4 {x}^{2} - 8 x + 12$

$y = - \left(0\right) {x}^{2} - 8 \left(0\right) + 12$

$y = 12$

as you can. see by what we, with did any quadratic function:

$y = a {x}^{2} + b x + c$

the y-intercept is just $c$.

graph{y=-4x^2-8x+12 [-20.71, 19.29, -3.68, 16.32]}