Question 188488
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Slight error in the wording of your problem.  Should have been:


"Find the vertex, x-intercept<i><b>s</b></i>, and y-intercept of the parabola given by *[tex \Large y=7x^2+21x-378]"


To find the vertex of a parabola whose equation is of the form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  f(x) = y = ax^2 + bx + c]


First find the <i>x</i>-coordinate of the vertex given by *[tex \Large x_V = -\frac{b}{2a} ]


Then find the <i>y</i>-coordinate by calculating the value of the function at the <i>x</i>-coordinate of the vertex:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \   y_V = f(-\frac{b}{2a}) = a(-\frac{b}{2a})^2 + b(-\frac{b}{2a}) + c]


To find the <i>x</i>-intercepts, i.e. the points where the value of the function, <i>y</i>, 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 <i>y</i>-intercept, i.e. the point where the graph intersects the <i>y</i>-axis -- and the point where the value of the independent variable is zero, substitute 0 for <i>x</i>.  For a quadratic function in standard form, the point is always (0,<i>c</i>).


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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