did i do this parabola question right?..

**Candy101** · December 15th 2009, 05:35 AM

i don't know if i did this right...

sketch 2x^2 +12x -14 and indicate all its vertex and intercepts

my vertex tht i got is (-3,-26)

my y-intercept is (0,-14) x-intercept (-1,0) and (7,0)
is this right?

**Stroodle** · December 15th 2009, 05:55 AM

Hi there.

One method of finding the minimum of the graph is to convert it to turning point form by completing the square:

$y=2x^2-12x-14$

$y=2(x^2+6x-7)$

$y=2((x+3)^2-16)$

$y=2(x+3)^2-32$

You can now see that the minimum is at $(-3,-32)$

Now to find the x-intercepts let $y=0$

$2(x+3)^2-32=0$

$(x+3)^2-16=0$

$(x+7)(x-1)=0$

Therefore the x-axis intercepts are at $(-7,0)$ and $(1,0)$

The y-intercept occurs where $x=0$ which is at $(0,-14)$

You should post your working so we can see where you went wrong.

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