00:01
This problem wants us to find a quadratic function f of x that has a minimum of negative 5 at x equals 3 and a y -intercept of negative 1.
00:09
So the important thing to remember about a quadratic is that we only have two types of shapes, either a right side up parabola or an upside down parabola, and if we're told that this quadratic function has a minimum, the only way we can make that happen for a quadratic is for our quadratic to open up.
00:26
So we know that that minimum will be our lowest point on this quadratic, which also means that it's the vertex.
00:33
So to begin building this function for f of x, we're going to start by looking at vertex form of a quadratic, which is a times x minus h squared plus k, where our h and k values are our vertex.
00:46
So if our minimum is negative 5 at x equals 3, that means 3 negative 5 is our minimum point, which means it's also our vertex.
00:54
So our h value will be 3 and our k value will be negative 5.
00:59
So at this point we have f of x equals our unknown value a times x minus 3 squared and then plus a negative 5, so minus 5 for k.
01:10
And it seems like we might not have enough information to figure out what a is, but we're also told that the y -intercept is supposed to be 1.
01:19
That means when we plug in the x value of 0, the result should be 1.
01:23
So we can use that to replace f of x with the y value 1 and show that's equal to a times the x value of 0 that gave us that y value of 1, still minus 3 squared minus 5...