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

2 Answers
Jun 20, 2018

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#

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=-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]}