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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
avatar+6179 
+1

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

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 Oct 2, 2019
 #2
avatar+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

 

 

cool cool cool

 Oct 2, 2019

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