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Mathematics 17 Online
OpenStudy (anonymous):

Question 4 (Multiple Choice Worth 1 points) Choose the equation below that corresponds to the graph shown. graph of a u-shaped figure opening downward with a maximum point of (−2, −2) y = −3x2 − 12x − 14 y = 3x2 + 12x − 14 y = −3x2 + 12x − 14 y = 3x2 − 12x − 14

OpenStudy (jdoe0001):

you know you can add a picture here by using the [Attach File] blue button

OpenStudy (anonymous):

OpenStudy (jdoe0001):

the vertex form of a parabola equation will be \(\bf y = a(x - h)^2 + k\) where the vertex of it is at (h , k) coordinates notice yours the vertex of it is at (-2, -2) that means (h, k) that means \(\bf y = a(x - h)^2 + k \implies y = a(x + 2)^2 -2\) so what's "a", well, we dunno but we can see another point it passes through, say (-1, -5) so we can say, that when x = -1, y = -5 so \(\bf y = a(x - h)^2 + k \implies y = a(x + 2)^2 -2\\ (-1, -5)\\ y = a(x + 2)^2 -2 \implies (-1) = a((-5) + 2)^2 -2\\ \implies -1 = a9-2 \implies 1 = 9a \implies \cfrac{1}{9} = a\)

OpenStudy (jdoe0001):

since the parabola is going downwards, so the "a" is negative, that is \(-\cfrac{1}{9} \)

OpenStudy (anonymous):

oh ok

OpenStudy (jdoe0001):

so, \(\bf y = a(x - h)^2 + k \implies y = a(x + 2)^2 -2\\ \implies y = -\cfrac{1}{9}(x + 2)^2 -2\) and we could expand that, to see what we get, for the "standard form"

OpenStudy (jdoe0001):

so..... see if you can expand it... and match it with one of your choices :)

OpenStudy (anonymous):

will it be this one y = 3x2 − 12x − 14

OpenStudy (jdoe0001):

dunno, you'd need to expand it to get the standard form

OpenStudy (anonymous):

I don't know that

OpenStudy (jdoe0001):

http://www.statisticslectures.com/images/trinomial11.gif <--- to expand the binomial

OpenStudy (anonymous):

ok but i don't know how to do that

OpenStudy (jdoe0001):

well.... if you were given the exercise, there seems to be an assumption that you're supposed to know how to, otherwise giving you the exercise would be improper

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