Question 200363
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Substitute 0 for *[tex \Large x] to calculate the *[tex \Large y]-intercept.  In both examples you provided it will be the value of the constant term in the function.  In general, the *[tex \Large y]-intercept for *[tex \Large ax^2 + bx + c] is *[tex \Large (0, c)]


Set the function equal to zero and solve for *[tex \Large x].  The *[tex \Large x]-intercept(s) will be the real number solution(s).  If the solution is a conjugate pair of complex numbers, then there are no *[tex \Large x]-intercepts.  This is the case with your first example.  The second example has a pair of real number intercepts.


In general if *[tex \Large b^2 - 4ac >= 0], then the *[tex \Large x]-intercepts of *[tex \Large ax^2 + bx + c] are *[tex \Large \left(\frac{-b \pm sqrt{b^2 - 4ac}}{2a},0\right)], otherwise there are no *[tex \Large x]-intercepts.


Functions of the form *[tex \Large ax^2 + bx + c], where *[tex \Large a \neq 0] graph to a parabola.  Parabolas have one turning point, namely the vertex.  The *[tex \Large x]-coordinate of the vertex of such a parabola is found by the calculation: *[tex \Large \frac{-b}{2a}].


The *[tex \Large y]-coordinate is the value of the function at that *[tex \Large x]-value, namely: *[tex \Large f\left(\frac{-b}{2a}\right) = a\left(\frac{-b}{2a}\right)^2 + b\left(\frac{-b}{2a}\right) + c]


Here are the graphs of your two examples.  First one in red, second one in green.


{{{drawing(
500, 500, -5, 5, -5, 5,
grid(1),
graph(
500, 500, -5, 5, -5, 5,
2x^2-8x+9,
x^2-x-2
))}}}

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