Question 399586
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ -2x^2\ -\ 4x\ +\ 15]


Standard form is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \rho(x)\ =\ a\left(x\ -\ h\right)^2\ +\ k]


Where *[tex \Large (h,k)] is the vertex.


Complete the square.  Factor the lead coefficient from the terms with variables:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ -2\left(x^2\ +\ 2x\right)\ +\ 15]


Divide the coefficient on the first degree term by 2, square the result, and add that result INSIDE the parentheses.  Compensate by adding the opposite of what you inserted inside the parentheses multiplied by the lead coefficient.

2 divided by 2 is 1, 1 squared is 1, add 1 inside the parentheses.  -2 times 1 is -2, the opposite of -2 is 2, add 2 to the end.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ -2\left(x^2\ +\ 2x\ +\ 1\right)\ +\ 15\ +\ 2]


Factor the trinomial inside the parentheses and collect terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ -2\left(x\ +\ 1\right)^2\ +\ 17]


Paying careful attention to the signs, by inspection, the vertex is at *[tex \Large (-1,17)]


The *[tex \Large x]-intercepts are the points where the graph of the function crosses the *[tex \Large x]-axis.  The *[tex \Large x]-coordinates of the *[tex \Large x]-intercepts are those values of *[tex \Large x] that make the value of the function equal zero.  Set the function equal to zero and solve.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -2\left(x\ +\ 1\right)^2\ +\ 17\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -2\left(x\ +\ 1\right)^2\ =\ -17]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\ +\ 1\right)^2\ =\ \frac{17}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 1\ =\ \pm\sqrt{\frac{17}{2}}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -1\ \pm\sqrt{\frac{17}{2}}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-2\ \pm\sqrt{34}}{2}]


So the *[tex \Large x]-intercepts are *[tex \Large \left(\frac{-2\ +\sqrt{34}}{2},0\right)] and *[tex \Large \left(\frac{-2\ -\sqrt{34}}{2},0\right)]


The *[tex \Large y]-intercept is the value of the function when *[tex \Large x] is zero.



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ -2\left(0\ +\ 1\right)^2\ +\ 17\ =\ 15]


So the *[tex \Large y]-intercept is *[tex \Large \left(0,15\right)] 



{{{drawing(
500, 500, -10,10,-20,20,
grid(1),
graph(
500, 500, -10,10,-20,20,
-2(x+1)^2+17))}}}


Don't know what you mean by "where do the numbers meet"  The Quadratic Club, or maybe at the Math Court at the Mall?  I give up.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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