Writing a quadratic when given the vertex and another point | Algebra 1 (TX TEKS) | Khan Academy

We're told a quadratic function f has a vertex at (-4, 7) and passes through the point (-2, -5). Write an equation for f in vertex form. So pause this video and try to work that out before we do that together.

All right, so first let's think about the general form of vertex form. If you have a quadratic function f, it's a function of x. So first, you're going to have some number times (x - h)^2 plus k. We'll talk in a little second what h and k are, but just to remind you why this right over here is actually a quadratic or a parabola, as you might recognize. You might say, "Okay, if I had something like this: f(x) = x^2," that makes sense—that that's going to have a shape something like that.

Now, you could also multiply that times a constant. So you could also imagine something like f(x) = a*x^2. If a is positive, you're still going to be upward opening. If a is negative, you would be downward opening like that. If a is greater than one, it would accelerate how fast as you go further and further from zero, how much that function increases. If it's between zero and one, it would kind of spread it out a little bit.

And then the rest of it is, well, imagine if you wanted to shift it to the right by h units. Well, then you would see something like this. If you wanted to shift it to the right by h units, you would have f(x) = a * (x - h)^2. That's just shifting a function. And then if you wanted to shift it up by k units, you would do + k. And that's exactly what we have right over here in this general form of a quadratic or quadratic in vertex form, I should actually say.

So the vertex is actually (h, k). This right over here, -4 is h and 7 is k. It's really telling you, if you didn't have h and k were zero, your vertex would be at zero. You would have just a traditional parabola, right, with the vertex at (0, 0). But we're shifting it in this form so that our vertex is at (-4, 7). So we can just substitute those in for h and k to start building out our the equation for f.

So, f(x) is going to be equal to a * (x - h), which is 4, we have to be very careful when we're subtracting negatives. So it's (-4) there squared plus k. k is 7.

Now, the next thing we can do—and why don't we just, well, the next thing we can do is simplify this a little bit. That uncertainty you heard in my voice was, well do I simplify this subtracting a negative and just making it adding four? I'll do that in the next step.

So the next thing I want to do is say, "Well, what is f(-2)?" Well, we know that f(-2) is -5, and we can use that to solve for a. So let me write this here. We could say f(-2), which we know is equal to -5, is equal to -5, but it's also going to be equal to all of this where I replace x with -2. So it's going to be equal to a times. So if I replace x with -2, it is actually—let me do it with—let me use that same color. -2 right over there, and then we have this subtracting a 4, so that's just adding a four squared, and then we have + 7.

And so now we just have to solve this part here. So we get -5 is equal to. What is -2 + 4? Well, that's just going to be 2.

2^2 is 4. So this all simplifies—let me make it clear—all this part right over here simplifies to 4a.

So we get: -5 = 4a + 7. We can just subtract seven from both sides and we get -12 = 4a. Divide both sides by four—oops!—divide both sides by four to solve for a, and we get a = -3.

So we’re actually done, but we want to write the whole equation out. We know what a is and we know what h and k are. So let's just write it out. The equation is f(x) = a, which we now know is -3, multiplied by (x + 4)^2 + 7.

And we are done.