Find a, b, and c

[mathdad]

The graph of the function f(x) = ax^2 + bx + c has a vertex at (1, 4) and passes through the point (-1, -8). Find a, b, and c.

I need the steps to solve this problem.

 

[Harry_the_cat]

Do you know the turning point (vertex) form of a quadratic function?

 

[mathdad]

Do you know the turning point (vertex) form of a quadratic function?

The vertex form of a quadratic function is given by f ( x ) = a ( x - h )^2 + k , where ( h, k ) is the vertex of the parabola. However, I requested the steps needed for me to solve on my own.

 

[Dr.Peterson]

The graph of the function f(x) = ax^2 + bx + c has a vertex at (1, 4) and passes through the point (-1, -8). Find a, b, and c.

I need the steps to solve this problem.

The vertex form of a quadratic function is given by f ( x ) = a ( x - h )^2 + k , where ( h, k ) is the vertex of the parabola. However, I requested the steps needed for me to solve on my own.

There are a couple different ways; if you are taking a course (or going through a book), it would be helpful to see what method fits into what you've learned, which is the reason for asking what you know.

One way is to complete the square, which puts it into vertex form. Evidently you don't have access to the steps for that, but you can find it in various places, such as the link in the previous sentence, or this one.

Another way is to use the formula [MATH]x = \frac{-b}{2a}[/MATH], which gives h in your formula, and which is the easy half of the quadratic formula, so you may already know it. That's found commonly on pages about graphing parabolas, like this one.

EDIT: Well, actually those are two ways to do the reverse of what you want to do. Sorry about that.

For your problem, you can just put h and k, which you were given, into the vertex form, and then all you need to do is to find a. One way to do that is to plug in not only h and k, but also x and y for the point you were given, leaving only a unknown. Then solve that equation for a. (And, of course, then expand what you have into the standard form.)

It would be a little less direct to use [MATH]x = \frac{-b}{2a}[/MATH] for this problem.

 

[Harry_the_cat]

The vertex form of a quadratic function is given by f ( x ) = a ( x - h )^2 + k , where ( h, k ) is the vertex of the parabola. However, I requested the steps needed for me to solve on my own.

As Dr P said, there are several ways to solve this problem. I asked if you knew the turning-point form because that would be my preferred starting point. If you didn't know it or hadn't seen it before, I would have suggested another way to approach the problem.

There is always a reason why we ask the questions we do.

You wouldn't be "solving it on your own" if we just gave you a list of possibly meaningless steps to follow.

 

[mathdad]

There are a couple different ways; if you are taking a course (or going through a book), it would be helpful to see what method fits into what you've learned, which is the reason for asking what you know.

One way is to complete the square, which puts it into vertex form. Evidently you don't have access to the steps for that, but you can find it in various places, such as the link in the previous sentence, or this one.

Another way is to use the formula [MATH]x = \frac{-b}{2a}[/MATH], which gives h in your formula, and which is the easy half of the quadratic formula, so you may already know it. That's found commonly on pages about graphing parabolas, like this one.

EDIT: Well, actually those are two ways to do the reverse of what you want to do. Sorry about that.

For your problem, you can just put h and k, which you were given, into the vertex form, and then all you need to do is to find a. One way to do that is to plug in not only h and k, but also x and y for the point you were given, leaving only a unknown. Then solve that equation for a. (And, of course, then expand what you have into the standard form.)

It would be a little less direct to use [MATH]x = \frac{-b}{2a}[/MATH] for this problem.

I will do as you suggested to find a. After finding a, how do I find b and c? Be advised that this question does not have a similar practice problem in the sample question pages.

Typically, the author(s) provide a set of sample questions at the start of each chapter or section and then similar practice problems at the end of the section or chapter. Keep in mind that about 95 percent of the questions that I decide to post here are not explained in the textbook(s). Obviously, if a question is explained and/or solved in the sample pages, then it's senseless for me to request guidance.

Agree? I will work on finding a given the vertex and point. I will show my work here, which I tend to do most of the time.

 

[mathdad]

As Dr P said, there are several ways to solve this problem. I asked if you knew the turning-point form because that would be my preferred starting point. If you didn't know it or hadn't seen it before, I would have suggested another way to approach the problem.

There is always a reason why we ask the questions we do.

You wouldn't be "solving it on your own" if we just gave you a list of possibly meaningless steps to follow.

Read my reply to Dr. Peterson.

 

[Harry_the_cat]

After you find "a" you will have the complete equation in vertex form. What do you get?

 

[mathdad]

f(x) = a(x - h)^2 + k

f(x) = a(x-1)^2+4

-8 = 4a+4

f(x) = -3(x-1)^2+4 = -3x^2+6x+1

a = -3, b = 6, c = 1

 

[Dr.Peterson]

Good work.

Some comments for anyone reading this:

This is a reminder that sometimes what looks like a mere (unhelpful?) question is really a major hint; you don't realize that until you try it out for yourself. Taking one step may make the next step easy to see.

Or it may not. Then you show what you found, and ask for the next hint.

As Harry suggested, one hint is worth a lot more than a list of steps, in helping you to think through the problem on your own. In life, you aren't generally given complete directions for solving a problem, but just a set of tools. That is probably why your book doesn't give complete examples to match every exercise. But, yes, sometimes it can be frustrating when you don't think of the right questions to ask yourself, and you need someone else at your side. That's why actual teachers can be valuable, as well as why sites like this can be good (but not when they just do the work for you).