find minimum or maximum of g(x)= -8x^2-38x-9

Question

 find minimum or maximum of g(x)= -8x^2-38x-9

Answer 1

Hi Jade.

 

To answer this, the easiest way is to graph g(x) = y=-8x² - 38x - 9.

 

You can go to Desmos online or any other online graphing calculuator or use a hand held one, but what you will find is that this looks like an upside-down "U", so it has a maximum, NOT a minimum.

 

To calculate it, first you need to know the x-coordinate of the max.  It is found by plugging in your a and b into the formula 

 

x = -(b/2a)

 

This is true for all parabolas.  Your a = -8.  Your b = -38

 

Plugging these in give the x-coordinate of the max = -(-38/-16) = -19/8 = -2.375

 

To find the value of the maximum, you now just need to plug this (-19/8) x-value into the equation for g(x) and get the corresponding y-value.  The y-value is the maximum.

 

I get (-8)(-2.375)² -38(-2.375) - 9 = (-8)(5.640625)+90.25-9

 

=-45.125+90.25-9 = 36.125 = 289/8

 

The max is 36.125 if you teacher will accept decimal point answers, or it is 289/8 if you need it as a fraction.

Answer 2

Since the leading coefficient of the parabola is negative, you will have a maximum.  Lets write the parabola in vertex form.

 

Use the formula

 

q = -b / (2a)  ---->  x-coordinate of maximum

 

where:

a = -8

b = -38

 

Then the maximum is the point (h, h(q)).

Answer 3

g(x) = -8x² - 38x - 9

g(x) = -8(x² + 19x/4) - 9

g(x) = -8(x² + 19x/4 + (19/8)²) - 9 - 361/8

g(x) = -8(x + 19/8)² - 433/8

 

Where is the vertex?