Question 188488: find the vertex, x-intercept, and y-intercept of the parabola given by y=7x^2+21x-378
Found 2 solutions by solver91311, stanbon:
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
Slight error in the wording of your problem. Should have been:
"Find the vertex, x-intercepts, and y-intercept of the parabola given by  "
To find the vertex of a parabola whose equation is of the form:
 = y = ax^2 + bx + c)
First find the x-coordinate of the vertex given by 
Then find the y-coordinate by calculating the value of the function at the x-coordinate of the vertex:
 = a(-\frac{b}{2a})^2 + b(-\frac{b}{2a}) + c)
To find the x-intercepts, i.e. the points where the value of the function, y, is zero, set the function equal to zero and solve the quadratic either by factoring, completing the square, or using the quadratic formula.
To find the y-intercept, i.e. the point where the graph intersects the y-axis -- and the point where the value of the independent variable is zero, substitute 0 for x. For a quadratic function in standard form, the point is always (0,c).
John
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! find the vertex, x-intercept, and y-intercept of the parabola given by
y = 7x^2+21x-378
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Vertex form:
7x^2 + 21x = y + 378
7(x^2 + 3x + (3/2)^2) = y + 378 + 7(9/4)
7(x+3/2))^2 = y + 393.75
Vertex: (-3/2 , -393.75)
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Y-intercept:
Let x= 0, then y = -378
-----------------------------------
x-intercept
Let y = 0 and solve for "x".
x = [-21 +- sqrt(21^2 - 4*7*-378)]/14
x = [-21 +- 105]/14
x = [-21 + 105]/14 or x = [-21 - 105]/14
x = 6 or x = -9
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Cheers,
Stan H.
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