Question

Consider the function f(x) = 10x² - 10x.

a. Determine, without graphing, whether the function has a minimum value or a maximum value.
b. Find the minimum or maximum value and determine where it occurs.
c. Identify the function's domain and its range.

Answer 1

Answers:

• A) The graph has a minimum value.
• B) The minimum value is f(x) = -2.5 and it occurs when x = 0.5
• C) Domain is "all real numbers"; range is
  `\boldsymbol{\text{y} \ge -2.5}`

The graph is shown below.

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Step-by-step explanation:

Part A)

x² is the same as x^2

The function f(x) = 10x^2-10x is the same as y = 10x^2-10x

Compare this to the template y = ax^2+bx+c to find the following

• a = 10
• b = -10
• c = 0

The value of 'a' is positive, which means the parabola opens upward. Furthermore, it tells us that the graph has a lowest point. There is no highest point since both parts of the parabola go upward forever.

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Part B)

We'll use the values of 'a' and b found earlier to get...

h = -b/(2a)

h = -(-10)/(2*10)

h = 10/20

h = 0.5

The min value therefore occurs when x = 0.5, which is the x coordinate of the vertex. The vertex being (h,k)

To find the value of k, we plug x = 0.5 into the original function.

y = 10x^2-10x

y = 10(0.5)^2-10*0.5

y = -2.5

This is the smallest the output y values can get. Therefore, it is the minimum value.

The vertex is located at (h,k) = (0.5, -2.5)

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Part C)

With any quadratic function, the domain is "set of all real numbers". We don't have to worry about things like...

• Division by zero errors.
• Taking the square root of negative values.

And so on.

The range is
`\text{y} \ge -2.5` since we found y = -2.5 to be the smallest y value possible. See part B above. So either y = -2.5 or y > -2.5

Check out the graph below. I used GeoGebra to make it, but you could use Desmos or other similar graphing tools if you prefer.