Question 189015
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  y = -x^2 + 2x + 3]


A parabola.


Opening downward because the lead coefficient is < 0.



<i>x</i>-coordinate of the vertex:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{-b}{2a} = \frac{-2}{2(-1)} = 1]


<i>y</i>-coordinate of the vertex:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  y = -(1)^2 + 2(1) + 3= -1 + 2 + 3 = 4]


Hence, the vertex is at


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (1, 4)]


Axis of symmetry:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{-b}{2a} = 1]


<i>y</i>-intercept:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  y = -(0)^2 + 2(0) + 3 = 3]


So <i>y</i>-intercept is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (0, 3)]


<i>x</i>-intercepts:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  -x^2 + 2x + 3 = 0]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \  (-x + 3)(x + 1) = 0]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = 3] or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = -1]


So the <i>x</i>-intercepts are


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (-1, 0)]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (3, 0)]


Since the <i>y</i>-intercept is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (0, 3)]


and the axis of symmetry is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = 1],


the point


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (2, 3)]


must also be on the graph.


{{{drawing(
500, 500, -5, 5, -5, 5,
grid(1),
blue(line(1, -5, 1, 5)),
green(circle(1,4,.1)),
green(circle(-1,0,.1)),
green(circle(3,0,.1)),
green(circle(0,3,.1)),
green(circle(2,3,.1)),
graph(
500, 500, -5, 5, -5, 5,
-x^2 + 2x + 3)
)}}}



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