+299 What is the Vertex, Y-intercept, X intercept, and the axis of symmetry of the equation?
y=-4x62+22x+12
+128408 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
