Question 316858
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -2x^2\ -\ 8x\ -\ 5]


Factor out -2 from the variable terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -2\left(x^2\ +\ 4x\right) -\ 5]


Divide the coefficient on the *[tex \Large x] term by 2 and square the result.  *[tex \Large \left(\frac{4}{2}\right)^2\ =\ 4]


Add 4 inside the parentheses and *[tex \Large -(-2)(4)\ =\ 8] outside the parentheses

 
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -2\left(x^2\ +\ 4x\ +\ 4\right) -\ 5\ +\ 8]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -2\left(x^2\ +\ 4x\ +\ 4\right)\ +\ 3]


Factor the perfect square in the parentheses:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -2\left(x\ +\ 2\right)^2\ +\ 3]


Vertex:  *[tex \LARGE \left(-2,3\right)]


Axis of symmetry:  *[tex \LARGE x\ =\ -2]


Domain: (All real numbers, just like any other polynomial function) *[tex \LARGE \{x\,|\,x\,\in\,\mathbb{R}\}]


Range:  The lead coefficient is negative, therefore the parabola opens downward and the vertex represents a maximum value of the function.  That maximum value is the *[tex \Large y]-coordinate of the vertex, namely *[tex \Large 3]. The range is unconstrained less than the maximum, so Range:  *[tex \LARGE \{y\, |\,y\,\in\,\mathbb{R},\ y\ \leq\ 3\}]


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You didn't ask for the intercepts, but here they are anyway:


The *[tex \Large y]-intercept is at the value of the function when *[tex \Large x\ =\ 0], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(0)\ =\ -2(0)^2\ -\ 8(0)\ -\ 5\ =\ -5]


and the *[tex \Large y]-intercept is *[tex \Large (0,-5)]


The *[tex \Large x]-intercepts are the roots of:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -2x^2\ -\ 8x\ -\ 5\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{4\ \pm\ \sqrt{6}}{-2}]


Verification of that last step is left as an exercise for the student.


Hence the *[tex \Large x]-intercepts are *[tex \Large \left(\frac{4\ +\ \sqrt{6}}{-2},0\right)] and *[tex \Large \left(\frac{4\ -\ \sqrt{6}}{-2},0\right)]


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