Math and Functions

Question

The parabola is f(x)=x²+2x-8. I need to find the relative maximum and the maximum value as well as the relative minimum and minimum value. If you could explain, that would be great. Thanks so much!

Answer 1

x²+2x-8

graphs as a parabola opening upward since the

coefficient of the squared term is positive.

That means the parabola has a minimum

value which is the output value, or y-value,

at the vertex.

 

The vertex will have coordinates: (-b/2a, f(-b/2a))

 

a = 1, b=2

 

-b/2a = -2/2 = -1

 

f(-1) = (-1)²+2(-1)-8

        = 1-2-8

        = -9

 

The vertex is at (-1,-9)

 

The minimum value is -9 at x=-1

 

There is no maximum since the graph continues up to infinity

on both sides of the vertex.

 

 

Answer 2

Relative minima and maxima depend on some interval, so I need some more info.  Sometimes these problems have the interval in bracket notation which can look like a coordinate pair when the interval is open, which is represented using parentheses.

 

The parabola you have presented opens upward, so its vertex is its absolute minimum.  The first term of the quadratic formula, negatvie b/2a, gives the x-coordinate of the vertex.  Solving the function gives the y-coordinate.

 

For your parabola in standard form:  x² + 2x - 8, a = 1 and b = 2.

 

The x-coordinate of the vertex is - 2/(2)(1) = - 1.

 

Solving the function for f(- 1) gives 1 - 2 - 8 = - 9

 

The vertex (absolute minimum) is at (- 1, - 9)

 

The parabola is unbounded above, which means that f(x) approaches infinity as x approaches infinity (or negative x approaches negative infinity).

 

An interval that does not include the vertex will have a relative minimum, which corresponds to a higher value of f(x) than that of the vertex.

 

The zeroes of the function are the x-intercepts.  The function (f(x)) is negative below a zero, and positive above.  An interval containing a zero of the function would have a relative minimum below the zero and a relative maximumu above.

 

For this parabola, solve for f(x) = 0

 

x² + 2x - 8 = 0

 

Attempt to factor using FOIL.

 

f(x) = (x + 4)(x - 2) = 0

 

The zeroes are at 2 and negative 4.

 

The closed interval [1, 3] would include the x-intercept 2.  The lowest point on this interval would be

 

f(1) = 1 + 2 - 8 = - 5

 

This is higher up the curve than the vertex at negative 9, as we expect.

 

Since the function is unbounded above, f(3), the highest point on the interval, is a relative maximum.

 

f(3) = 9 + 2(3) - 8 = 7, which is positive, as we expect.