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avatar+290 

What is the Vertex, Y-intercept, X intercept, and the axis of symmetry of the equation?

 

y=-4x62+22x+12

 Mar 14, 2019
 #1
avatar+98061 
+3

I think you might have

 

y = -4x^2 + 22x + 12

 

In the form     Ax^2 + Bx + C.....the x coordinate of vertex is given by  -B / [2A]

So...here the x coordinate of the vertex is  -22/ {2 * - 4)  =  -22/-8  =  11/4  = 2.75

 

And to find the y coordinate we have   -4(11/4)^2 + 22(11/4) + 12   = 42.25

 

So....the vertex is (2.75, 42.25 )

 

y intercept.....let x = 0   and we have   -4(0)^2 + 22(0) + 12  =    12

 

x  intercept ......let y = 0  and solve for x

 

0 = - 4x^2 + 22x + 12      divide through by - 2

 

0 = 2x^2 - 11x - 6       factor

 

(2x + 1 ) ( x -6)  = 0      set  each factor to 0 and solve for x  and we have  x = -1/2  and x = 6

 

And those are the x intercepts

 

The axis of symmetry  is just    x =  the x cooordinate of the vertex =  2.75

 

 

cool cool cool

 
 Mar 14, 2019
 #2
avatar+290 
+2

Thank you so much CPhill !

 
 Mar 14, 2019
 #3
avatar+98061 
0

OK.....

 

 

cool cool cool

 
CPhill  Mar 14, 2019

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