SOLUTION: how can we know if the graph of quadratic equation opens? a. upward b. downward c. to the left d. to the right

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Question 219414: how can we know if the graph of quadratic equation opens?
a. upward
b. downward
c. to the left
d. to the right
Found 2 solutions by jsmallt9, Theo:
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
The basic forms of quadratic equations are:
y+=+ax%5E2+%2B+bx+%2B+c
and
x+=+ay%5E2+%2B+by+%2B+c

The first form, y+=+ax%5E2+%2B+bx+%2B+c, with an x%5E2 term (not y%5E2) opens upward or downward. If "a", the coefficient of x%5E2, is positive the parabola opens upward. And if "a" is negative the parabola opens downward.

The second form, x+=+ay%5E2+%2B+by+%2B+c, with an y%5E2 term (not x%5E2) opens to the right or to the left. If "a", the coefficient of y%5E2, is positive the parabola opens to the right. And if "a" is negative the parabola opens to the left.

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
standard form of a quadratic equation is y = f(x) = ax^2 + bx + c if it opens up or down.
standard form of a quadratic equation is x = f(y) = ay^2 + by + c if it opens left or right.
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If your equation is in the form of y = ax^2 + bx + c:
It opens up if a is positive.
It opens down if a is negative.
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If your equation is in the form of x = ay^2 + by + c
It opens to the right if a is positive.
It opens to the left if a is negative.
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If it's in the form of x = f(y) = ay^2 + by + c (open left or right), you might also see it in the alternate form of:
y = f(x) = %28-b%2F2a%29 +/- sqrt%28%28%28x-c%29%2Fa%29%2B%28b%2F%282a%29%29%5E2%29
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You take the x = f(y) version and solve for y to get the y = f(x) version.
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Once again, though, if a is positive it opens to the right and if a is negative it opens to the left.
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To see this with real equations, consider:
y = x^2 + 2x + 1
This opens up as shown in the following graph because the a term of 1 is positive:
graph%28300%2C300%2C-5%2C5%2C-5%2C5%2Cx%5E2+%2B+2x+%2B+1%29
Now consider:
y = -x^2 + 2x + 1
This opens down as shown in the following graph because the a term of -1 is negative:
graph%28300%2C300%2C-5%2C5%2C-5%2C5%2C-x%5E2+%2B+2x+%2B+1%29
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Now consider:
x = y^2 + 2y + 1
Before we can graph this, we have to solve for y to get:
y = -1 +/- sqrt(x)
This opens to the right as shown in the following graph because the coefficient of x under the square root sign is positive. It is positive because the a term was positive:
graph%28300%2C300%2C-5%2C5%2C-5%2C5%2C-1%2Bsqrt%28x%29%2C-1-sqrt%28x%29%29
We converted x = f(y) to y = f(x) in order to graph the equation. This is the alternate form i was talking about.
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Now consider:
x = -y^2 + 2y + 1
Before we can graph this, we have to solve for y to get:
y = 1 +/- sqrt(-x+2)
This opens to the left as shown in the following graph because the coefficient of x under the square root sign is negative. It is negative because the a term was negative:
graph%28300%2C300%2C-5%2C5%2C-5%2C5%2C1-sqrt%28-x%2B2%29%2C1%2Bsqrt%28-x%2B2%29%29
We converted x = f(y) to y = f(x) in order to graph the equation. This is the alternate form I was talking about.
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