# Graphing Parabolas =/

## Homework Statement

For problems 1 and 2 determine if the quadratic opens up or down. Find the vertex. Find any intercepts and use these things to graph the quadratic equation.

1. y=-3x^2+12x-9

2. y=2x^2+3x+2

## The Attempt at a Solution

For number 1:
ok well i determined the parabola opens down becuase a(-3) is negitive. Hope i got that right.
Then i found the vertex by h=-b/2a becuase that will give me the x cordinate of the vertex so i did -12/2(-3) = 2. 2 is the x cordinate of the vertex so i put 2 into the original equation to get the y cordinate
Y=-3(2)^2+12(2)+9=21 i get 21 as my Y Cordinate.

Now for the intercepts i had a friend help me with this and he said there are no y int for a parabola...he also to me to take the equation -3x^2+12x-9=0 set it to 0 like that.

Then he told me to divid by 3 and get x^2-4x+3=0 then factor to get x=3 and x= 1 and those are my x ints for the parabola is this correct? any help would be greaty apriciated

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Y=-3(2)^2+12(2)+9=21 i get 21 as my Y Cordinate.
Looks good so far except that at this stage you wrote +9, but in the original equation you wrote -9.

Now for the intercepts i had a friend help me with this and he said there are no y int for a parabola...
That isn't true. The y intercept is the value of y=-3x2+12x-9 when x = 0. To find it, just set x equal to 0, and solve for y.

he also to me to take the equation -3x^2+12x-9=0 set it to 0 like that. Then he told me to divid by 3 and get x^2-4x+3=0 then factor to get x=3 and x= 1 and those are my x ints for the parabola is this correct? any help would be greaty apriciated
That's right. (Another method is to use the quadratic formula.)