Use Descartes’s rule of signs to determine the possible number of positive and negative real zeros of f(x)=8x^4-6x^3-4x^2-7x+3
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Answer:
a). Vertex form f the function f(x) = (x - 2.5)²+3.25
b). Maximum value of the function is (-3)
Step-by-step explanation:
a). f(x) = x² - 5x + 3
we have to write this function in the vertex form.
f(x) = x² - 2(2.5)x + 3
= [(1)²x²- 2(2.5)x + (2.5)²-(2.5)²] + 3
= (x - 2.5)²- 6.25 + 3
= (x - 2.5)² - 3.25
So the vertex for of the function is
f(x) = (x - 2.5)² - 3.25
b). Given quadratic function f(x) = -x² - 6x + 6 is in the standard form of f(x) = ax² + bx + c.
By comparing the coefficients of both the functions,
a = -1, b = 6 and c = 6
Here the value of a is negative, therefore, the given parabola is opening downwards.
And the maximum will occur at
When we plug in the values,
x = 3
Now to get the value of maximum of the function we will get the value of f(3).
f(3) = -(3)²- 6(3) + 6
f(3) = 9 - 18 + 6
f(3) = -3
Therefore, maximum value of the quadratic function is (-3).