Writing a quadratic when given the vertex and another point
About this video
This video writes a quadratic in vertex form, f(x) = a(x - h)^2 + k, given its vertex (-4, 7) and another point (-2, -5). It explains what a, h and k do to the graph, substitutes the vertex for h and k, then uses the extra point to solve for a, giving f(x) = -3(x + 4)^2 + 7.
You'll be able to
- explain how a, h and k in f(x) = a(x - h)^2 + k change the graph of a parabola.
- substitute a vertex (h, k) into vertex form, taking care with negative values.
- use another point on the parabola to solve for a and write the full equation.
Key words
- Vertex
- The turning point of a parabola, its highest or lowest point.
- Vertex form
- A way of writing a quadratic as f(x) = a(x - h)^2 + k, where (h, k) is the vertex.
- Parabola
- The U-shaped graph of a quadratic function.
- Upward opening
- A parabola shaped like a U, which happens when a is positive.
- Downward opening
- A parabola shaped like an upside-down U, which happens when a is negative.
Transcript
- [Instructor] We're told a quadratic function f has a vertex at negative four, seven and passes through the point negative two, negative five. 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.
So if you have a quadratic function f, it's a function of x. So first, you're going to have some number x minus h squared plus k, and we'll talk in a little second what a, 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 of x is equal to x squared, 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 of x is equal to ax squared. 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 of x is equal to a times x minus h squared.
That's just shifting a function. And then if you wanted to shift it up by k units, you would do a plus k. And that's exactly what we have right over here in this general form of a quadratic or a quadratic in vertex form, I should actually say. So the vertex is actually h and k specify the vertex.
This right over here, negative four is h and seven is k. It's really telling you if you didn't have, if h and k were zero, your vertex would be at zero, you would have just a traditional parabola, right with the vertex at zero. But we're shifting it in this form so that our vertex is at x equals negative four and y is equal to seven.
So we can just substitute those in for h and k to start building out the equation for f. So f of x is going to be equal to a times x minus h is negative four. We have to be very careful when we're subtracting negatives. So it's negative four there squared plus k, k is seven.
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 wanna do is say, well, what is f of negative two? Well, we know that f of negative two is negative five and we can use that to solve for a. So let me write this here. So we could say f of, I'll do that same color, f of negative two, which we know is equal to negative five is equal to negative five, but it's also going to be equal to all of this where I replace x with a negative two.
So it's going to be equal to a times. So if I replace x with a negative two, it is, actually, let me use that same color, negative two right over there. And then we have this subtracting a negative four. So that's just adding a four, squared and then we have plus seven.
And so now we just have to solve this part here. So we get negative five is equal to... What is negative two plus four? Well, that's just going to be two. Two squared is four. So this all simplifies. Lemme make it clear, all this part right over here simplifies to 4a is equal to 4a plus seven, plus seven.
We can just subtract seven from both sides and we get negative 12 is equal to 4a. Divide both sides by four. Whoops. Divide both sides by four to solve for a. And we get a is equal to negative 12 divided by four, which is equal to negative three. Negative three is equal to A and we're actually done, but we wanna 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 of x is equal to A, which we now know is negative three times x minus negative four, which we could just write that as x plus four. Make that x a little bit cleaner. X plus four squared plus seven, plus seven and we are done.
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