The "x intercept" is where y = 0, which is where the curve crosses the x axis.
Solve y = x^2+ 2x- 8 = 0
You can write that as
y = (x+4) (x-2) = 0
There are two places where y = 0. The factored equation tells you where they are.
how would I find the x-intercepts of
y=x^2+2x-8
and
y=x^2-5x-10
thank you, I'm really having trouble getting this.
y = x^2-5x-10 does not factor easily.
If you solve x^2-5x-10 = 0 with the quadratic equation, you get
y = [5 +/- sqrt 65]/2
= +6.531 and -1.531
To find the x-intercepts of a quadratic equation, you need to solve for the values of x when y is equal to zero.
Let's start with the equation y = x^2 + 2x - 8. To find the x-intercepts, we need to solve this equation when y is equal to zero.
We can rewrite the equation as (x + 4) (x - 2) = 0. This is because the equation is already factored for us. From this factored form, we can see that there are two possible solutions for x that would make the equation equal to zero: x = -4 and x = 2.
Therefore, the x-intercepts for the equation y = x^2 + 2x - 8 are x = -4 and x = 2.
Now let's move on to the equation y = x^2 - 5x - 10. This equation does not factor easily, so we will have to use the quadratic formula to find the x-intercepts.
The quadratic formula is given by x = [-b ± sqrt(b^2 - 4ac)] / (2a), where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c = 0.
For our equation y = x^2 - 5x - 10, the coefficients are: a = 1, b = -5, and c = -10.
Plugging these values into the formula, we get:
x = [-(-5) ± sqrt((-5)^2 - 4(1)(-10))] / (2(1))
x = [5 ± sqrt(25 + 40)] / 2
x = [5 ± sqrt(65)] / 2
So the x-intercepts for the equation y = x^2 - 5x - 10 are approximately x = 6.531 and x = -1.531.
In summary:
- For the equation y = x^2 + 2x - 8, the x-intercepts are x = -4 and x = 2.
- For the equation y = x^2 - 5x - 10, the x-intercepts are approximately x = 6.531 and x = -1.531.