Question

\(h(x)=-2 x^{2}+x\) (a) Find the average rate of change from 0 to 3 . (b) Find an equation of the secant line containing \((0, h(0))\) and \((3, h(3))\)

Short answer

a) -5. b) y = -5x.

Step-by-step solution

- Find the value of h(0) and h(3)

First, calculate the values of the function at the endpoints of the interval. For this, we substitute the values back into the function: \(h(0) = -2(0)^2 + 0 = 0\) \(h(3) = -2(3)^2 + 3 = -18 + 3 = -15\)

- Calculate the average rate of change

The average rate of change of a function between two points is given by: \(\frac{h(3) - h(0)}{3 - 0}\). Substituting the values from Step 1: \(\frac{-15 - 0}{3 - 0} = \frac{-15}{3} = -5\)

- Find the slope of the secant line

The slope of the secant line between \((0, h(0))\) and \((3, h(3))\) is the same as the average rate of change calculated in Step 2, which is -5.

- Use point-slope form to find the equation of the secant line

The point-slope form of a line is \(y - y_1 = m(x - x_1)\), where m is the slope and \((x_1, y_1)\) is a point on the line. Using \(m = -5\), \(x_1 = 0\), and \(y_1 = 0\): \(y - 0 = -5(x - 0)\) Or simplified: \(y = -5x\)

Key concepts

Average Rate of Change

The average rate of change of a function between two points helps us understand how the function's values change over that interval. In this case, we have the function given by: \(h(x) = -2x^2 + x\).
To find the average rate of change from \(x = 0\) to \(x = 3\), we first calculate the values of the function at these points.
Substituting \(x = 0\) into the function, \(h(0) = -2(0)^2 + 0 = 0\).
Next, for \(x = 3\), we have \(h(3) = -2(3)^2 + 3 = -18 + 3 = -15\).
Now, using the formula for the average rate of change, we get:
\(\frac{h(3) - h(0)}{3 - 0} = \frac{-15 - 0}{3 - 0} = \frac{-15}{3} = -5\).
This tells us that, on average, the function decreases by 5 units for every 1 unit increase in \(x\) between \(0\) and \(3\).

Point-Slope Form

The point-slope form of a line equation is crucial for finding the equation of a line when we know a point on the line and the slope.
The general formula for point-slope form is:
\(y - y_1 = m(x - x_1)\),
where \(m\) represents the slope and \((x_1, y_1)\) is a point on the line.
In our exercise, the slope \(m\) we found is -5, and our points are \((0, h(0)) = (0, 0)\) and \((3, h(3)) = (3, -15)\).
Let's use \((0, 0)\) and \(m = -5\) in the point-slope form:
\(y - 0 = -5(x - 0)\).
Simplifying, we get the equation of the secant line as:
\(y = -5x\).

Quadratic Function

A quadratic function is a type of polynomial function that has the general form:
\(f(x) = ax^2 + bx + c\), where \(a eq 0\).
These functions create parabolas when graphed. The exercise given to us has: \(h(x) = -2x^2 + x\),
where the coefficient \(a = -2\), and \(b = 1\), and \(c = 0\).
The parabola opens downward since the coefficient of \(x^2\) is negative.
Understanding the shape and properties of quadratic functions can help us in finding their critical points, intersections, and understanding their rate of change.

Slope Calculation

The slope of a line measures its steepness and direction. It is calculated as the ratio of the change in the vertical direction (rise) to the change in the horizontal direction (run), often described simply as 'rise over run'.
Mathematically, it is given by:
\(m = \frac{y_2 - y_1}{x_2 - x_1}\).
In our exercise, the slope of the secant line between the points \((0, 0)\) and \((3, -15)\) was found by:
\(m = \frac{-15 - 0}{3 - 0} = \frac{-15}{3} = -5\).
This means that for every 1 unit increase in \(x\), the function \(h(x)\) decreases by 5 units.
Knowing how to calculate the slope is essential for analyzing linear relationships and understanding how two points on a graph are connected.