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MTH163
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http://vhcc2.vhcc.edu/pc1fall9/properties_of_quadratics/properties_of_quadratics.htm
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I believe I am struggling to understand specific quadratics. I am looking at :
Example 1: y = x^2 - 3x - 4.
As an example, consider the graph of the function y = x^2 - 3x - 4.
The vertex of the graph is at xVertex = - b / (2a) = - (-3) / (2 * 1) = 3/2.
***** here I get lost, I am assuming the formula is vertex = -b / (2a) , in this example where are you getting your numbers 2 and 3???? *****
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3 is the coefficient of x.
2 comes from the expression (2 a). a being the coefficient of x^2, which is 1, the expression (2a) gives us 2.
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The corresponding y coordinate is found by substituting x = 3/2 into the definition of the function:
yVertex = (3/2)^2 - 3(3/2) - 4 = 9/4 - 9/2 - 4 = 9/4 - 18/4 - 16/4 = -25/4.
Thus the vertex is at (3/2, -25/4) or, in decimal form, (1.5,-6.25).
The graph crosses the x axis when y is zero. By the statement of the quadratic formula, this occurs when x^2 - 3x - 4 = 0. This equation is in the form specified by the formula, with a = 1, b = -3 and c = -4. By the statement of the formula, the zero value is at
x = [ - (-3) +- `sqrt( (-3)^2 - 4(1)(-4) ) ] / (2 * 1) = [ 3 +- `sqrt(25) ] / 2 = [3 +- 5] / 2.
The two solutions are therefore
x = [3 + 5] / 2 = 8/2 = 4
and
x = [3 - 5] / 2 = -2 / 2 = -1.
So the graph crosses the x axis at x = 4 and at x = -1.
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