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.