Help me find a formula for a quadratic function.

wilc0919

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Oct 13, 2006
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Suppose that a quadratic function has its vertex at (3,0) and has a y-intercept of -4. FInd a formula for this function.

I missed this calss...and have no idea where to start....can someone explain the answer so I can use it to answer other questions?
 
You have the vertex form of the quadratic:

. . . . .y = a(x - h)<sup>2</sup> + k

Plug in the given values for the vertex. Then plug in the given point, and solve for "a". Then re-write the equation, with the values given for h and k, and the value you just found for "a".

Eliz.
 
plug what where?

how do I do that?

is it

0=a(3-h)^2 + k

what am I solving for there? and where do I and how do I plug in the y intercept?
I am totally lost.
 
wilc0919 said:
how do I do that?
Do which? Are you not familiar with quadratics or parabolas, so you don't know what the vertex is, or which number is "h" and which is "k"...?

(If you've covered this material in class and/or in your book, you should know this, is why I ask.) Since you don't seem to be familiar with the notation or terms, it sounds like you are requesting lessons on these topics. Since we cannot teach courses within this environment, it would be helpful if you specified which parts of this material have not been covered yet.

Thank you.

Eliz.
 
wilc0919 said:
Suppose that a quadratic function has its vertex at (3,0) and has a y-intercept of -4. FInd a formula for this function.

I missed this calss...and have no idea where to start....can someone explain the answer so I can use it to answer other questions?

If a parabola has a vertex at (h, k) and a vertical axis of symmetry (opens up or down), its equation can be written in this form (which is called vertex form):

y = a(x - h)<SUP>2</SUP> + k

The vertex is (h, k).....so "h" is the x-coordinate of the vertex, and "k" is the y-coordinate of the vertex.

Your parabola has its vertex at (3, 0), so h = 3 and k = 0. Substititute into the general form of the equation:

y = a(x - h)<SUP>2</SUP> + k
y = a(x - 3)<SUP>2</SUP> + 0

or,
y = a(x - 3)<SUP>2</SUP>

You may note that we still don't know what the value of "a" is, and we need that to complete the equation. We DO know that the y-intercept of the parabola is -4....and this is the same as saying the parabola passes through the point (0, -4). So we have a point which must satisfy the equation. That is, the equation must be true if we substitute 0 for x and -4 for y. Let's do that:

y = a(x - 3)<SUP>2</SUP>
-4 = a(0 - 3)<SUP>2</SUP>

Ok...now you've got an equation in which the only variable is "a". You can solve this for a. Substitute the value you get for a into y = a(x - 3)<SUP>2</SUP> to complete the equation you're asked for.

I hope this helps you.
 
is this right?

FIRST THANKS SO MUCH FOR THE HELP.

Here's what I got when trying to write a formula for the function when a quadratic equation has its vertex at (3,0) and a y-intercept of -4.

y=a(x-3)^2 + 0
is
y=a(x-3)^2

then to solve for a I get
-4=a(0-3)^2
is
a= -4/9

so am I right when I say my answer is
y= -4/9(x-3)^2
 
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