How can I find the X-Intercept of this equation?

Question

I had to put this equation into standard form of a parabola: f(x)=(-x^2+6x-8). The standard form is: -(x-3)^2 +1. I found that the Vertex is (3,1), and the Yint was (0,-8) because I set all the X's as 0 because I was trying to find Y. But how do I do it for the Xint? If there is only one Y? y=-(x-3)^2 -1

Sorry if I am missing it... I just don't get it.

Answer 1

For figuring the y-intercept, you set *x* to zero and solve for y.
So to figure the x-intercept(s), you set *y* to zero and solve for x.

Starting with the equation for your parabola:
-x² + 6x -8 = 0

You can multiply both sides by -1, which will give you a plain x².
x² -6x + 8 = 0

Now factor:
(x - 2)(x - 4) = 0

So the x intercepts are x = 2 and x = 4. The points are (2,0) and (4,0)

You can see that in the graph below:

Answer 2

The x-intercepts are the solutions to the equation f(x)=0, so if f(x) is a quadratic function, then there are at most 2 of them.

You can compute them using the quadratic formula, or, if you put the equation into standard form (by completing the square), you can find them fairly easily.

To solve
-(x-3)^2 + 1 = 0,
isolate the squared term:
(x-3)^2 = 1.
Take the square root:
x-3 = 1 or x-3 = -1
Solve for x:
x=4 or x=2.

Therefore, the x-intercepts are (4,0) and (2,0).

Answer 3

Set y = 0 and solve for x.
0 = -x^2 + 6x - 8
Multiply both sides by -1
0 = x^2 - 6x + 8
Factor
0 = (x-2)(x-4)
x = 2 or x = 4
The x-intercepts are (2,0) and (4,0)

Answer 4

If you are trying to figure out x-int, you need to set y = 0.

0 = -(x-3)^2+1
0 = -x^2 + 6x - 9 + 1
0 = x^2 - 6x + 8
0 = (x - 4)(x - 2)
x = 4, 2

x-int: (4, 0) and (2, 0)