Find the Vertex of the Parabola 3x^2 + 18x - 3

[Monique316]

Please help me with the steps. This is what I have so far:

f(x) = 3x^2 + 18x - 3

f(0) = 3(0)^2 + 18(0) - 3

f(0) = -3

So the Y intercept is (0,-3)?? Then I got lost trying to solve the rest of the problem.

Please help. Thank you.

 

[Mrspi]

Re: Find the Vertex of the Parabola.

Please help me with the steps. This is what I have so far:

f(x) = 3x^2 + 18x - 3
f(0) = 3(0)^2 + 18(0) - 3
f(0) = -3
So the Y intercept is (0,-3)?? Then I got lost trying to solve the rest of the problem.

Please help. Thank you.

If by "the rest of the problem," you mean finding the vertex, there are several ways you might do that.

Here's one:

When the equation of a parabola is in the form

f(x) = ax<SUP>2</SUP> + bx + c

the x-coordinate of the vertex is -b/2a.

Your equation is in this form, with a = 3 and b = 18. So, the x-coordinate of the vertex is -18/(2*3), or -3.

To find the y-coordinate of the vertex evaluate f(x) for x = -3.

I hope this helps you.

 

[Monique316]

I'm sorry if I'm a little slow with this, but can you please write out the entire problem exactly how you would solve it?

I still don't understand.

Thank you.

 

[stapel]

...can you please write out the entire problem exactly how you would solve it?

Most (legitimate) tutors don't "do" students' work for them, so it would probably be wise to show at least an attempt at following the instructions you were provided earlier.

The assignment gave you a quadratic in the form f(x) = ax² + bx + c. The tutor gave you a formula, in terms of a and b, for finding the x-coordinate of the vertex, and then did the computations for you. Then the tutor gave you the instructions for finding the y-coordinate of the vertex.

Please show some effert in doing this one last step. Thank you.

Eliz.

 

[Mrspi]

I'm sorry if I'm a little slow with this, but can you please write out the entire problem exactly how you would solve it?

I still don't understand.

Thank you.

f(x) = ax<SUP>2</SUP> + bx + c
The x-coordinate of the vertex is x = -b/2a

f(x) = 3x<SUP>2</SUP> + 18x + 3
The x-coordinate of the vertex is x = -18 / (2*3)
x = -18/6
x = -3

To find the y-coordinate of the vertex, find f(x) when x = -3
f(x) = 3x<SUP>2</SUP> + 18x + 3
f(-3) = 3(-3)<SUP>2</SUP> + 18(-3) + 3
f(-3) = 3(9) - 54 + 3
f(-3) = 27 - 54 + 3
f(-3) = -24

The vertex is at (-3, -24)

If you can't follow that, then I suggest you have a heart-to-heart talk with your academic advisor. It is possible that you have been placed in a class you're not prepared for.

 

[Monique316]

I greatly appreciate you showing me the correct process. I did the same thing, unfortunatley, for some strange reason, I came up with -22.

Thank you. and I will go back over my steps.