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Fed [463]
3 years ago
15

Find the minimum value of the function f(x) = x2 + 5x – 6

Mathematics
1 answer:
Klio2033 [76]3 years ago
6 0

So the easiest method to find the vertex (the minimum in this case) to do this is to find the axis of symmetry, then plug it into the function.

Firstly, the equation to find the axis of symmetry is x=\frac{-b}{2a} , with b = x coefficient and a = x^2 coefficient. The equation equation can be solved as such:

x=\frac{-5}{2*1}\\\\x=\frac{-5}{2}\\\\x=-2.5

Since the vertex falls on the axis of symmetry, we know that the x-coordinate of the vertex is -2.5. Now to solve for the y-coordinate, plug in x with -2.5 and solve as such:

f(-2.5)=(-2.5)^2+5*(-2.5)-6\\ f(-2.5)=6.25+(-12.5)-6\\ f(-2.5)=-12.25

Now putting it all together, our minimum value (vertex) is (-2.5,-12.25).

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Answer:

a) (0.555, 0) and (6, 0)

b) r = -3 and r = 1.8

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Step-by-step explanation:

Set each term in the numerator and denominator equal to 0 and find r.

In the numerator:

r = 7/8, 5/9, or 6

In the denominator:

r = 9/5, 7/8, or -3

Zeros in the numerator that aren't in the denominator are r-intercepts.

Zeros in the denominator that aren't in the numerator are vertical asymptotes.

Zeros in both the numerator and the denominator are holes.

a) (0.555, 0) and (6, 0)

b) r = -3 and r = 1.8

c) Evaluate m(r) at r = 7/8.  To do that, first divide out the common term (-8r + 7) from the numerator and denominator.

m(r) = (-9r+5)² (r−6)² / ( (-5r+9)² (r+3)² )

m(⅞) = (-9×⅞+5)² (⅞−6)² / ( (-5×⅞+9)² (⅞+3)² )

m(⅞) = (-23/8)² (-41/8)² / ( (37/8)² (31/8)² )

m(⅞) = (-23)² (-41)² / ( (37)² (31)² )

m(⅞) = 0.676

The hole is at (0.875, 0.676).

d) Evaluate m(r) at r = 0.

m(0) = (-9×0+5)² (0−6)² / ( (-5×0+9)² (0+3)² )

m(0) = (5)² (-6)² / ( (9)² (3)² )

m(0) = 1.235

The m(r)-intercept is (0, 1.235).

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The ball will hit the ground at 0.125 seconds

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The equation of the function is a quadratic function, and the function is given as:

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So, the equation of the function becomes

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Divide both sides by -2t

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Hence, the ball will hit the ground at 0.125 seconds

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