Question 235130
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First you want to graph the function *[tex \LARGE f(x)\ =\ y\ =\ x^2\ -\ 2x\ -\ 24]


Then you want to find the one place that the graph of the function is tangent to the *[tex \Large x]-axis, meaning that you have 1 real solution (actually, you would say that the equation has one real root with a multiplicity of two), or you want to find the two places that the graph intersects the *[tex \Large x]-axis, meaning that you have two distinct real solutions, or you want to discover that the graph never touches the *[tex \Large x]-axis, meaning that the equation has no real solutions.


What you are doing is answering the question, "What number or numbers can I substitute for *[tex \Large x] so that *[tex \Large f(x)\ =\ 0]?"  The function has a value of zero at the *[tex \Large x]-axis.


{{{drawing(
400, 600, -10, 10,-30, 30,
grid(1),
graph(
400, 600, -10, 10,-30, 30,
x^2-2x-24)
)}}}


Compare this to: *[tex \LARGE f(x)\ =\ y\ =\ x^2\ -\ 2x\ +\ 1]


{{{drawing(
400, 600, -10, 10,-30, 30,
grid(1),
graph(
400, 600, -10, 10,-30, 30,
x^2-2x+1)
)}}}


Notice two things:  1.  The function is a perfect square trinomial.  2.  The graph is tangent to the axis.  One root with a multiplicity of 2.


No compare to: *[tex \LARGE f(x)\ =\ y\ =\ x^2\ -\ 2x\ +\ 4]


{{{drawing(
400, 600, -10, 10,-30, 30,
grid(1),
graph(
400, 600, -10, 10,-30, 30,
x^2-2x+4)
)}}}


Notice that the graph does not touch the *[tex \Large x]-axis at all.  No real solutions.






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