# How do you find the vertex and the intercepts for #y=2x^2-12x#?

y =0 at x=0

Divide both sides by 2

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To find the vertex of the quadratic function ( y = 2x^2 - 12x ), you can use the formula for the x-coordinate of the vertex: ( x = \frac{-b}{2a} ), where ( a ) is the coefficient of the ( x^2 ) term and ( b ) is the coefficient of the ( x ) term.

For the given function, ( a = 2 ) and ( b = -12 ). Substituting these values into the formula, you get:

[ x = \frac{-(-12)}{2(2)} = \frac{12}{4} = 3 ]

To find the y-coordinate of the vertex, substitute ( x = 3 ) into the function:

[ y = 2(3)^2 - 12(3) = 2(9) - 36 = 18 - 36 = -18 ]

So, the vertex of the parabola is ( (3, -18) ).

To find the x-intercepts, set ( y = 0 ) and solve for ( x ):

[ 2x^2 - 12x = 0 ] [ 2x(x - 6) = 0 ]

This equation is satisfied when either ( 2x = 0 ) or ( x - 6 = 0 ).

So, the x-intercepts are ( x = 0 ) and ( x = 6 ).

To find the y-intercept, substitute ( x = 0 ) into the function:

[ y = 2(0)^2 - 12(0) = 0 ]

So, the y-intercept is ( (0, 0) ).

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