Test 1 Section 4 #30

To get this one, it’s helpful (but not necessary, if you’re nimble enough) to know the vertex form of a parabola, which is y=a(x-h)^2+k where (h,k) is the vertex of the parabola. If you know that, then you can recognize immediately that only choice D is even in that form! And, sure enough, choice D says y=(x-1)^2-16, which indicates a vertex at (1,-16). That’s consistent with the given figure.

Even if you don’t know the vertex form, though, you can get this if you read the question and figure carefully, as long as you know that the vertex of a parabola is its minimum or maximum. You need the coordinates of the vertex to show up in the equation as constants. Just by looking at the figure you can see that the x-coordinate is definitely 1, even if you’re not sure what the y-coordinate is because of the scale of the y-axis. Choice D is the only choice with a 1 in it as a constant! (You’re also in the calculator allowed section, so you are totally allowed to graph the original equation and find the exact vertex if you don’t trust your eyeballs).

But OK, we can do the algebra, too. To solve this algebraically, we need to convert from the given equation to the vertex form (above) by completing the square.


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