xintercepts

[reebok35]

How do you find the xintercepts for the graph of the quadratic relation y=xsquared-2x-35?

 

[Mrspi]

How do you find the xintercepts for the graph of the quadratic relation y=xsquared-2x-35?

The x-intercepts are the point(s) where the graph of the equation crosses the x-axis. You need to remember that any point ON the x-axis has a y-coordinate of 0. So, if you are looking for x-intercepts, substitute 0 for y.

Here's an example of a problem similar to yours:

Find the x-intercepts (if any) for the graph of y = x[sup:3eym7uk6]2[/sup:3eym7uk6] + 4x - 12

Since I know that an x-intercept will be of the form (x, 0), I'll substitute 0 for y:

0 = x[sup:3eym7uk6]2[/sup:3eym7uk6] + 4x - 12

Now I need to solve the resulting quadratic equation, which I could do with the quadratic formula, or perhaps by factoring.

The right side factors rather easily, so I'll do it that way:

0 = (x + 6)(x - 2)

Now, set each factor equal to 0 and solve.

If (x + 6)(x - 2) = 0, then x + 6 = 0, or x - 2 = 0.
x + 6 = 0
x = -6

x - 2 = 0
x = 2

So, the x-intercepts of the graph are at (-6, 0) and (2, 0)

Try the same method on your problem!

 

[reebok35]

How would u do it for y=-2(x+5)squared+18?

 

[Mrspi]

How would u do it for y=-2(x+5)squared+18?

Again, if you are looking for x-intercepts, you need to substitute 0 for y and solve the resulting equation.

Here's an example that is similar to your problem:

Find the x-intercepts (if any) for the equation y = 3(x + 2)[sup:bfvz3ixk]2[/sup:bfvz3ixk] - 27

Since any point on the x-axis has a y-coordinate of 0, the x-intercept(s) will be of the form (x, 0). Substitute 0 for y:

y = 3(x + 2)[sup:bfvz3ixk]2[/sup:bfvz3ixk] - 27

0 = 3(x + 2)[sup:bfvz3ixk]2[/sup:bfvz3ixk] - 27

Add 27 to both sides to isolate the term containing x:

0 + 27 = 3(x + 2)[sup:bfvz3ixk]2[/sup:bfvz3ixk] - 27 + 27

27 = 3(x + 2)[sup:bfvz3ixk]2[/sup:bfvz3ixk]

Divide both sides by 3:

27/3 = [3(x + 2)[sup:bfvz3ixk]2[/sup:bfvz3ixk]) / 3

9 = (x + 2)[sup:bfvz3ixk]2[/sup:bfvz3ixk]

Take the square root of both sides of the equation. Remember that any positive number has TWO square roots; one positive and one negative.

+ sqrt(9) = sqrt[(x + 2)[sup:bfvz3ixk]2[/sup:bfvz3ixk]]

+ 3 = x + 2

so, 3 = x + 2 OR -3 = x + 2

Solve each of those equations:

1 = x OR -5 = x

The x-intercepts are (1, 0) and (-5, 0)

An alternative way to solve is this:

0 = 3(x + 2)[sup:bfvz3ixk]2[/sup:bfvz3ixk] - 27
Expand the right side by doing the squaring:
0 = 3(x[sup:bfvz3ixk]2[/sup:bfvz3ixk] + 4x + 4] - 27
Now simplify the right side:
0 = 3x[sup:bfvz3ixk]2[/sup:bfvz3ixk] + 12x + 12 - 27
0 = 3x[sup:bfvz3ixk]2[/sup:bfvz3ixk] + 12x - 15
And either factor the right side, or use the quadratic formula. You should get the same values of x that I did using the method I showed.