Problem 4 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Sketch the graph of each of these functions. Be sure to show all intercepts and any high or low points. a. \(f(x)=3 x-5\) b. \(f(x)=-x^{2}+3 x+4\)

Step-by-Step Solution

Answer

For part (a), plot (0, -5) and \( \left( \frac{5}{3}, 0 \right)\) and draw a line. For part (b), plot (0, 4), (-1, 0), (4, 0) and \( \left( \frac{3}{2}, \frac{17}{4} \right)\) and draw a downward parabola.

Step 1: Determine the type of function for part (a)

The function is a linear equation, given by \(f(x)=3x-5\). Linear functions graph to straight lines.

Step 2: Find the y-intercept and x-intercept for part (a)

To find the y-intercept, set \(x=0\): \(f(0)=3(0)-5 = -5\). The y-intercept is (0, -5). For the x-intercept, set \(f(x)=0\): \(0=3x-5\). Solving for \(x\), \(x= \frac{5}{3}\). The x-intercept is \(\left(\frac{5}{3}, 0\right)\).

Step 3: Graph the linear function for part (a)

Plot both intercepts: (0, -5) and \(\left( \frac{5}{3}, 0 \right)\). Draw a straight line through these points.

Step 4: Determine the type of function for part (b)

The function is a quadratic equation, given by \(f(x)=-x^{2}+3x+4\). Quadratic functions graph to parabolas.

Step 5: Find the y-intercept for part (b)

Set \(x=0\) in the equation: \(f(0)=-0^2+3(0)+4=4\). The y-intercept is (0, 4).

Step 6: Find the x-intercepts for part (b)

Set \(f(x)=0\): \(-x^2+3x+4=0\). Solving this quadratic equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) where \(a=-1\), \(b=3\), and \(c=4\), we get \(x = \frac{-3 \pm 5}{-2}\). This results in two solutions: \(x=-1\) and \(x=4\). So the x-intercepts are (-1, 0) and (4, 0).

Step 7: Find the vertex for part (b)

The vertex of a parabola given by \(f(x)=ax^2+bx+c\) occurs at \(x=\frac{-b}{2a}\). Substituting \(a=-1\) and \(b=3\), we get \(x=\frac{-3}{-2}=\frac{3}{2}\). Substitute \(x=\frac{3}{2}\) back into the function to find the y-coordinate: \(f\left( \frac{3}{2} \right) = -\left( \frac{3}{2} \right)^2 + 3 \left( \frac{3}{2} \right) + 4 = \frac{17}{4}\). The vertex is at \(\left( \frac{3}{2}, \frac{17}{4} \right)\).

Step 8: Graph the quadratic function for part (b)

Plot the y-intercept (0, 4), x-intercepts (-1, 0) and (4, 0), and the vertex \(\left( \frac{3}{2}, \frac{17}{4} \right)\). Sketch the parabola opening downwards since the coefficient of \(x^2\) is negative.

Key Concepts

linear functions

Linear functions are one of the simplest types of functions to understand and graph. The general form of a linear function is given by: \( f(x) = mx + b \). Here, 'm' represents the slope of the line and 'b' represents the y-intercept, where the line crosses the y-axis.

To graph a linear function, you only need two points. These points are often the x-intercept and y-intercept.
- The x-intercept is found by setting \( f(x) = 0 \) and solving for 'x'.
- The y-intercept is found by setting \( x = 0 \) and solving for 'f(x)'.
Once you have these two points, you plot them on the graph and draw a straight line through them.

For example, consider the function \( f(x) = 3x - 5 \).
- The y-intercept is found by setting \( x = 0 \): \( f(0) = -5 \). So the y-intercept is (0, -5).
- The x-intercept is found by setting \( f(x) = 0 \): \( 0 = 3x - 5 \) which gives \( x = \frac{5}{3} \). Thus, the x-intercept is \( \left( \frac{5}{3}, 0 \right) \).

Once plotted, a straight line through these points will give the graph of the linear function.

quadratic functions

Quadratic functions graph to parabolas and have the general form of: \( f(x) = ax^2 + bx + c \), where 'a', 'b', and 'c' are constants.

The value of 'a' determines the direction of the parabola:
- If 'a' is positive, the parabola opens upwards.
- If 'a' is negative, the parabola opens downwards.

The graph of a quadratic function features a vertex, which is its highest or lowest point, and it might intersect the x-axis at two points (real & distinct roots), one point (a repeated root), or not at all (complex roots).

To accurately graph a quadratic function, you need:
- The y-intercept which is found by setting \( x = 0 \)
- The x-intercepts which are the roots of the equation \( f(x) = 0 \)
- The vertex, found using \( x = \frac{-b}{2a} \) and substituting back into the equation to get the y-value.

For instance, consider \( f(x) = -x^2 + 3x + 4 \). As 'a' is negative, this parabola opens downward. With the y-intercept (0, 4) and solving the quadratic equation, the x-intercepts are (-1, 0) and (4, 0).

function intercepts

Function intercepts are where the graph crosses the axes:

- The x-intercept(s) of a function is/are the point(s) where the function value (output) is zero. To determine x-intercepts, set \( f(x) = 0 \) and solve for 'x'. This will result in the points \( (x, 0) \).
- The y-intercept of a function is the point where the input is zero. To determine the y-intercept, set \( x = 0 \) and solve for 'f(x)' which results in the point \( (0, f(0)) \).

For example, in the linear function \( f(x) = 3x - 5 \), setting \( x = 0 \) gives us the y-intercept (0, -5). Setting \( f(x) = 0 \) and solving for x gives \( x = \frac{5}{3} \), so the x-intercept is \( \left( \frac{5}{3}, 0 \right) \).

In the quadratic function \( f(x) = -x^2 + 3x + 4 \), the y-intercept is (0, 4), found by setting \( x = 0 \). To find the x-intercepts, solve \( -x^2 + 3x + 4 = 0 \), which gives us the intercepts (-1, 0) and (4, 0).

parabola vertex

The vertex of a parabola is the point where it reaches its maximum or minimum value, depending on the direction it opens.

In a quadratic function \( f(x) = ax^2 + bx + c \), the vertex can be determined using the formula \( x = \frac{-b}{2a} \).
- Once the x-coordinate of the vertex is found, substitute it back into the original equation to find the y-coordinate.

For instance, in the quadratic function \( f(x) = -x^2 + 3x + 4 \), the x-coordinate of the vertex is found with \( a = -1 \) and \( b = 3 \), yielding \( x = \frac{-3}{-2} = \frac{3}{2} \). Substituting \( x = \frac{3}{2} \) back into the function, the y-coordinate is \( f \left( \frac{3}{2} \right) = \frac{17}{4} \). Therefore, the vertex of this parabola is \( \left( \frac{3}{2}, \frac{17}{4} \right) \).

The vertex is a crucial point in graphing parabolas, as it gives a clear direction of the parabola's curvature, either opening upwards or downwards.