How do you find the vertex and intercepts for $y=2x^2 +11x-6$ ?

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Answer 1 · 2017-09-14T12:47:13

Vertex is at $(-2.75, -21.125)$ x intercepts are at
$(-6,0) and (0.5,0)$ , y intercept is at $(0, -6)$

$y= 2x^2+11x-6 or y= 2(x^2+11/2x)-6$

Adding $121/8$ on both sides we get ,

$y= 2{x^2+11/2x+(11/4)^2] -121/8 -6$ or

$y= 2(x+11/4)^2 -169/8$ or

$y= 2(x+2.75)^2 -21.125$ . Comparing with standard vertex form

of equation $y=a(x-h)^2+k ; (h,k)$ being vertex , we find here

$h= -2.75 , k = -21.125 , a=2$ . So vertex is at $(h,k)$ or

$(-2.75, -21.125)$ , y intercept can be found by putting

$x=0$ in the equation as $y = 2*0 +11*0 -6$ or

$y = -6 or (0,-6)$ , x intercept can be found by putting $y=0$

in the equation as $0 = 2x^2 +11x -6 or 2x^2 +12x -x -6=0$

or $2x(x+6) -1 (x+6) =0 or (2x-1)(x+6)=0$

$:. x = 0.5 or x= -6$ , x intercepts are at $(-6,0) and (0.5,0)$

graph{2x^2+11x-6 [-80, 80, -40, 40]} [Ans]

How do you find the vertex and intercepts for y=2x^2 +11x-6y=2x^2 +11x-6?