graphing parabola for y = x^2 - 3x - 10

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jhawk555

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Sep 26, 2006
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I ran into a snag and was looking for some guidance on the following equation.

. . .y = x^2 - 3x - 10

The leading coefficient a is greater than 0, so the parabola opens up.

. . .a = 1
. . .b = -3
. . .c = -10

The formula for "h" from the vertex gives me:

. . .-b/(2a) = -(-3)/(2(1)) = 3/2

So the vertex is at: (h, k) = (3/2, 31/4)

I don't think I did something right.

. . .x-intercepts: (5, 0) and (-2, 0)
. . .y-intercept: (0, -10)

Did I do this correctly?
 
Re: graphing parabolas

jhawk555 said:
I ran into a snag and was looking for some guidance on the following equation.

. . .y = x^2 - 3x - 10

The leading coefficient a is greater than 0, so the parabola opens up.

. . .a = 1
. . .b = -3
. . .c = -10

The formula for "h" from the vertex gives me:

. . .-b/(2a) = -(-3)/(2(1)) = 3/2

So the vertex is at: (h, k) = (3/2, 31/4)

I don't think I did something right.

. . .x-intercepts: (5, 0) and (-2, 0) Correct
. . .y-intercept: (0, -10) Correct

Did I do this correctly?
Click to expand...
Your vertex needs some work: your x-coordinate ("h") is correct, but try replugging it in your original equation to find the y-coordinate ("k"):

. . .(3/2)^2 -3(3/2) -10

. . . . .=(9/4) - (9/2) -10=???

Other than that, good work!
 
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