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

y=-4x62+22x+12

Mar 14, 2019

#1
+107098
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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

Mar 14, 2019
#2
+299
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Thank you so much CPhill !

Mar 14, 2019
#3
+107098
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OK.....

CPhill  Mar 14, 2019