`y=2x^2+4x-3` Find the vertex and intercepts.  

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mlehuzzah | Student, Graduate | (Level 1) Associate Educator

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`y=2x^2+4x-3`

To find the vertex, use the formula `x=-b/(2a)`

Here we have `x=-4/(2*2) = -1`

To find the y-value of the vertex, plug in the x value:

`y=2(-1)^2+4(-1)-3=-5`

So the vertex is (-1,-5)




To find the y-intercept, you plug in x=0 and solve for y.

`y=2(0)^2+4(0)-3=-3`

So the y-intercept is (0,-3)




To find the x-intercepts, you plug in y=0 and solve for x. This is a little harder:

`0=2x^2+4x-3`

There are two ways to solve something like this. You can factor it, or, if that doesn't work, you can use the quadratic formula:

`x = -b/(2a) +- (sqrt(b^2-4ac))/(2a)`

`x= -4/(2*2) +- (sqrt(16-(-24)))/(2*2) = -1 +- (sqrt(40))/4`

`x=.58` or `x= -2.58`





The graph is

 







PS I edited your question. If you really did mean `y<2x^2+4x-3`, the graph is the same, but with a dotted line instead of a solid one, and then shaded below the parabola.

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mlehuzzah | Student, Graduate | (Level 1) Associate Educator

Posted on

PPS I forgot to mention it - the vertex is the spot on the graph where the parabola "turns around"

It is the very bottom of the parabola (or, if the parabola opens down, it is the very top of the parabola)

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