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# help

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Find the quadratic function yequals​f(x) whose graph has a vertex ​(negative 3​,3​) and passes through the point ​(negative 6​,0). Write the function in standard form.

Oct 2, 2019

#1
+6179
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$$\text{From the vertex we have}\\ (y-3) = a(x+3)^2\\ \text{Now from the point}\\ (0-3) = a(-6+3)^2\\ -3 = 9a\\ a=-\dfrac 1 3\\ y = -\dfrac 1 3 (x+3)^2 + 3$$

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Oct 2, 2019
#2
+109450
+2

Thanks, Rom.....

In standard form we have

y = (-1/3) (x+ 3)^2 + 3

y = (-1/3) (x^2 + 6x +9) + 3

y = (-1/3)x^2 - 2x - 3  + 3

y = (1/3)x^2 - 2x

Oct 2, 2019