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maks197457 [2]
3 years ago
7

Let f(x) be a polynomial function such that f(−4)=−5,f′(−4)=0 and f′′(−4)=1. classify the point (−4,−5).

Mathematics
1 answer:
svet-max [94.6K]3 years ago
3 0
THis will be  the coordinates of the point  at a turning point on the graph.

Since the second derivative  is positive (=1)  it will be a minimum value of f(x).
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Use the parabola tool to graph the quadratic function y=-1? - 2.0 +8
Andreas93 [3]

Answer:

see explanation

Step-by-step explanation:

I don't have graphing facilities but can give you the vertex and 1 other point.

Given a parabola in standard form

y = ax² + bx + c ( a ≠ 0 )

Then the x- coordinate of the vertex is

x = - \frac{b}{2a}

y = - x² - 2x + 8 ← is in standard form

with a = - 1 and b = - 2 , then

x = - \frac{-2}{-2} = - 1

Substitute x = - 1 into the equation for corresponding value of y

y = - (- 1)² - 2(- 1) + 8 = - 1 + 2 + 8 = 9

vertex = (- 1, 9 )

To obtain another point substitute any value for x into the equation

x = 0 : y = 0 - 0 + 8 , then (0, 8 ) is a point on the graph

x = 2 : y = - (2)² - 2(2) + 8 = - 4 - 4 + 8 = 0  then (2, 0 ) is a point on the graph

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