Question

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

Answer 1

The horizontal intercepts have coordinates `(1,0)` and `(-3,0)`

Explanation:

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

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

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

`0=-4x^2-8x+12`

or, if you prefer,

`4x^2+8x-12=0`

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

`x^2+2x-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`

Answer 2

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=-4x^2-8x+12`

`0=-4x^2-8x+12`

Factor:

`0=-4 (x + 3)(x - 1)`

`x=-3` and `x=1`

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

`y=-4x^2-8x+12`

`y=-(0)x^2-8(0)+12`

`y=12`

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

`y=ax^2+bx+c`

the y-intercept is just `c`.

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