Writing transformation form equations of polynomials from a graph
About this video
The video finds the equation of a quartic graph in transformation form, f(x) = a(x - h)^4 + k. Starting from y = x^4, it flips the graph to open down, shifts it left 1 and down 2, giving f(x) = -(x + 1)^4 - 2, and checks the answer using the point (0, -3).
You'll be able to
- describe what a, h and k do in f(x) = a(x - h)^n + k.
- write the transformation form of a quartic from its graph.
- check an equation by testing a point on the graph.
Key words
- transformation form
- Writing a function as a(x - h)^n + k so the scaling and shifts of a simpler function are easy to see.
- vertex
- The turning point of a graph like y = x^4, the highest or lowest point.
- fourth degree polynomial
- A polynomial whose highest power of x is 4.
- opens down
- A graph whose ends both point downward, like y = -x^4.
Transcript
- [Instructor] So the goal of this video is to find the equation for this function right over here, this function f, and they want us to write it in transformation form. And so just as a review, transformation form would be like this, f of x is equal to a times x minus, and we're assuming that if we were to expand this out, it would be some form of a polynomial, so x minus h to the nth power plus k.
This is considered transformation form because you could imagine a simpler function, f of x is equal to x to the n. And then if you shift it to, let's say, well, if you shift it right or left by h, you might change this value over here. If you shift it up or down, that's what k is gonna determine.
And if you scale it, that's what a is gonna determine. So if you transform this by shifting to the left or up or down or scaling it, you'd get something like f of x or that original f of x. Now, the other hint I will give you is I'll tell you that if you were to expand this out, if we were to expand the equation out here, right over here, this would be a fourth degree polynomial.
So we immediately know that the form of f of x is a times x minus h to the fourth power plus k. So let's think about how we would approach it. So one way to approach it is, first, let's just think about what y equals x to the fourth would look like. Well, y equals x to the fourth, when x is zero, y would be zero.
When x is one, y would be one. When x is negative one, y would be one. when x is 0.5 or is 1/2, 1/2 to the fourth is 1/16, so it's gonna be really low, it's gonna look something like that, and same thing when we're at negative 1/2. So the graph is going to look something like this.
I'm not gonna graph the whole thing, but you get the idea. Well, we immediately see a problem. This is opening up while our f of x opens down. So let's try to make it open down. So let's think about y is equal to negative x to the fourth. Well, that by the same logic is going to just flip that red one over.
So it's going to look like this. So this is already going to be interesting. Now, it looks like we just have to shift this to the left by one and then down by two if we just think about where this vertex is. So if you shift to the left by one, what we're going to do is we're gonna change the x, so we have negative, instead of x to the fourth, if you shift to the left by one, so if we shift to the left right over here, you would replace the x with x plus one.
So y is equal to negative x plus one to the fourth is going to look something like this. And then last but not least, if we're going to shift that down by two to get to our f of x, we just need to subtract two here. Our k would be negative two. So we would get f of x is equal to negative x plus one to the fourth minus two.
And we could even validate that this point is indeed on that curve, if we were to say f of zero, that's negative, and then I'll say zero plus one to the fourth power minus two. So this is going to be equal to, this is all going to be equal to one, so this is equal to negative one minus two, which is equal to negative three.
F of zero is indeed equal to negative three. And we're done.
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