How do I find the x intercepts of the parabola y=x^2-6x-7?

Thanks

The x intercepts are where y = 0. There are two of them.

Note that y = x^2 -6x -7 = (x-7)(x+1),
by factoring.

y is zero when either of those factors is zero.

The y intercept is where x = 0. The equation tells you that happens at y = -7.

The X-intercepts are the points where the parabola crosses the X-axis when

Y=0.The x-intercepts are also the
solutions to the equation:
Y=X^2-6X-7=0,solve for X by factoring,
(X+1)(X-7)=0, X+1=0,X=-1; X-7=0,X=7.
The X-INT(solutions)are:X=-1, X=7.

To find the x-intercepts of a parabola, you need to set the y-coordinate equal to zero and solve for x.

Given the equation y = x^2 - 6x - 7, we can set y equal to zero and solve for x:

0 = x^2 - 6x - 7

To solve this quadratic equation, we can either factor it or use the quadratic formula. In this case, factoring is not possible, so we'll use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In our equation, a = 1, b = -6, and c = -7. Substituting these values into the quadratic formula, we get:

x = (-(-6) ± √((-6)^2 - 4(1)(-7))) / (2(1))

Simplifying further:

x = (6 ± √(36 + 28)) / 2
x = (6 ± √64) / 2
x = (6 ± 8) / 2

Now, we have two possible solutions for x:

1. x = (6 + 8) / 2 = 14 / 2 = 7
2. x = (6 - 8) / 2 = -2 / 2 = -1

So, the x-intercepts of the parabola y = x^2 - 6x - 7 are x = 7 and x = -1.

To find the x-intercepts of a parabola, you need to find the values of x when y equals zero. In other words, you are looking for the points where the parabola crosses the x-axis.

To find the x-intercepts of the given parabola y = x^2 - 6x - 7, you need to set y = 0 and solve for x.

So, let's set y = 0:

0 = x^2 - 6x - 7

Now, you can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:

The quadratic formula states that for a quadratic equation of the form ax^2 + bx + c = 0, the solutions for x can be found using the formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In our case, a = 1, b = -6, and c = -7. Substituting these values into the quadratic formula, we get:

x = (6 ± √((-6)^2 - 4(1)(-7))) / (2(1))

Simplifying further, we have:

x = (6 ± √(36 + 28)) / 2

x = (6 ± √64) / 2

x = (6 ± 8) / 2

Now, we have two possibilities:

For the positive root:

x = (6 + 8)/2
x = 14/2
x = 7

For the negative root:

x = (6 - 8)/2
x = -2/2
x = -1

Therefore, the parabola intersects the x-axis at the points (7, 0) and (-1, 0). These are the x-intercepts of the given quadratic equation.