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Numerade Educator

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Problem 10 Medium Difficulty

Find expressions for the quadratic functions whose graphs are shown.

Answer

$$
\begin{array}{c}{f(x)=2(x-3)^{2}} \\ {\text { and }} \\ {g(x)=-x^{2}-2.5 x+1}\end{array}
$$

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Video Transcript

all right, so in this problem we have two different quadratic functions graft, and we're going to find their equations, and I'm going to do each part a little bit differently. So for the first part, what I'm going to do is start with the most basic or essential quadratic function Y equals X squared and think about its features. So we know that it would have its Vertex at 00 and we know it would go through the 0.11 So it goes over one up, one from the Vertex so we could slide that over. We could shift it over to the right three units, and we would get the graph y equals X minus three quantity squared. What that would look like is the same graph, but shifted three units to the right, so it's Vertex would be a 30 And instead of going through 11 since we shifted it three units to the right, it would go through for one. Now, let's take a look at the one we're trying to figure out. It has a vertex of 30 just like the one I just drew. But instead of going through 41 It goes through four to that tells us there was a vertical stretch and it's twice as tall as the one I just drew. So that means we need to multiply our equation by two. And that will give us why equals two times the quantity X minus three squared as the equation of this parabola. Now let's take a look at the second part. We're gonna do this one a little bit differently were given three points that fall on the parabola. And we know the standard form for the equation of ah, quadratic is y equals X squared plus B x Fix that a little bit Why equals a X squared plus BX plus c Well, weaken Dio is our quest here is to find the values of A B and C and we have three different points. We can substitute into this equation. So when we substitute and the 0.1 we get one equals zero times x squared plus zero times x plus c That tells us that C is one. Okay, we found one of the three values we need. We can also substitute the point negative 22 into the equation and we have two equals eight times negative two squared plus B times negative too. Class C. And we already know that C is one. So let's simplify this. This would be two equals four a minus to B plus one. If we subtract one from both sides, we have one equals for a minus to B. Now let's go through that process with the third point, the 0.0.1 negative 2.5. So if we substitute negative 2.5 in for why and one in for X, we get negative. 2.5 equals a times one squared plus B times one plus c and we know that C is one. So let's simplify that and we have negative 2.5 equals a plus B plus one. And if we subtract one from both sides, we get negative. 3.5 equals a plus B. So we have two equations with two unknowns. We have a system we can solve for A and B. Okay, let's go ahead and make some more space to do that. So these were the two equations that we just had, and what I'm going to do is multiply the second equation by two and then added to the first equation. So we have negative seven equals two a plus to be We have these equations together. This is the elimination method for solving a system. We get negative. Six equals six a divide both sides by six and we have a equals negative one. Now let's go ahead and find B and we'll have all three values. So if we substitute that number and to one of the equations will have in this case, negative 3.5 equals negative one plus B Add one to both sides and we have negative 2.5 equals b. Okay, so we have C c was one. We have a and we have be, Let's put it all together. Why equals negative one times X squared, minus 2.5 times X plus one. And that's the equation for that particular quadratic