Let f(x) be a polynomial function such that f(−4)=−5,f′(−4)=0 and f′′(−4)=1. classify the point (−4,−5).
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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 = -
y = - x² - 2x + 8 ← is in standard form
with a = - 1 and b = - 2 , then
x = - = - 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