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The parabola y=ax^2+k has a vertex (0,-4) and passes through the point (3,7). Find the equation.

 Apr 6, 2020
 #1
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Find a formula for the quadratic function whose graph has its vertex at (1,4) and its y-intercept at y-6.

 Apr 6, 2020
 #2
avatar+111329 
+2

y =  ax^2   +  k

 

If  the  vertexx is  (0, -4),  then  k   = -4

 

And  the  point (3,7)  is on the parbola implies that

 

7 =  a (3)^2  -  4     add 4  to  both  sides

 

11  = a (9)    divide both sides  by  9

 

11/9  = a

 

So.....the  equation is

 

y = (11/9)x^2  -  4

 

 

cool cool cool 

 Apr 6, 2020
 #3
avatar+932 
+1

This is a correction nobody will care about but this question is in Algebra 1 not precalc.

indecision

 Apr 6, 2020

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