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What is the vertex of #y= -6x^2 + 4x + 3#?

Answer 1

Vertex: #(1/3, 3 2/3)#

Probably the easiest way to do this is to convert the equation into "vertex form": #y=m(x-a)^2+b# with vertex at #(a,b)#

Given: #color(white)("XXX")y=-6x^2+4x+3#

Extract the #m# factor #color(white)("XXX")y= (-6)(x^2-2/3x)+3#

Complete the square #color(white)("XXX")y=(-6)(x^2-2/3x+(1/3)^2)+3-(-6)*(1/3)^2#

Rewrite with a squared binomial and simplified constant #color(white)("XXX")y=(-6)(x-1/3)^2+ 3 2/3#

which is in vertex form with vertex at #(1/3, 3 2/3)#

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Answer 2

Eli Assouline

To find the vertex of the quadratic function y = -6x^2 + 4x + 3, use the formula for the x-coordinate of the vertex: x = -b/(2a), where a = -6 and b = 4. Then, plug this x-coordinate into the equation to find the y-coordinate.

x-coordinate of the vertex: x = -4/(2*(-6)) = -4/(-12) = 1/3.

Now, plug x = 1/3 into the equation to find the y-coordinate:

y = -6(1/3)^2 + 4(1/3) + 3 = -6(1/9) + 4/3 + 3 = -2/3 + 4/3 + 3 = 3 - 2/3 = 9/3 - 2/3 = 7/3.

So, the vertex of the quadratic function is (1/3, 7/3).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.