Problem 15 - Precalculus Mathematics for Calculus

Question

A quadratic function is given. (a) Express the quadratic function in standard form. (b) Find its vertex and its \(x\) - and \(y\) -intercept(s). (c) Sketch its graph. $$f(x)=-x^{2}+6 x+4$$

Step-by-Step Solution

Answer

The vertex is (3, 13), x-intercepts are approximately (6.6, 0) and (-0.6, 0), and the y-intercept is (0, 4).

Step 1: Express the quadratic function in standard form

The standard form of a quadratic function is \( f(x) = ax^2 + bx + c \). The given function is \( f(x) = -x^2 + 6x + 4 \), which is already in standard form with \( a = -1 \), \( b = 6 \), and \( c = 4 \).

Step 2: Find the vertex

The vertex of a quadratic function in standard form \( ax^2 + bx + c \) is given by \( \left( \frac{-b}{2a}, f\left( \frac{-b}{2a} \right) \right) \). Substitute the values of \( a = -1 \) and \( b = 6 \): \( x = \frac{-6}{2(-1)} = 3 \). Substitute \( x = 3 \) into the function to find \( y \): \( f(3) = -(3)^2 + 6(3) + 4 = -9 + 18 + 4 = 13 \). So, the vertex is \( (3, 13) \).

Step 3: Find the x-intercepts

To find the \( x \)-intercepts, we need to solve \( -x^2 + 6x + 4 = 0 \). Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = -1 \), \( b = 6 \), and \( c = 4 \): \( x = \frac{-6 \pm \sqrt{36 - 4(-1)(4)}}{-2} = \frac{-6 \pm \sqrt{52}}{-2} \). \( \sqrt{52} = 2\sqrt{13} \), so \( x = \frac{-6 \pm 2\sqrt{13}}{-2} = 3 \pm \sqrt{13} \). The \( x \)-intercepts are \( 3 + \sqrt{13} \) and \( 3 - \sqrt{13} \).

Step 4: Find the y-intercept

The \( y \)-intercept is found by evaluating the function at \( x = 0 \). Substitute \( x = 0 \) into the function: \( f(0) = -0^2 + 6(0) + 4 = 4 \). Hence, the \( y \)-intercept is \( (0, 4) \).

Step 5: Sketch the graph

To sketch the graph, plot the vertex \((3, 13)\), \( x \)-intercepts \((3 + \sqrt{13}, 0)\) and \((3 - \sqrt{13}, 0)\), and the \( y \)-intercept \((0, 4)\). The parabola is downward opening because \( a = -1 < 0 \). Draw a symmetrical curve passing through these points, opening downwards.

Key Concepts

Vertex of a Quadratic Function

The vertex of a quadratic function is a crucial point that helps us understand the graph's shape and behavior. A quadratic function is typically expressed in the form \( f(x) = ax^2 + bx + c \). The vertex represents the highest or lowest point of the graph, known as the parabola, depending on the direction it opens.
For the function \( f(x) = -x^2 + 6x + 4 \), the values of \( a \), \( b \), and \( c \) are -1, 6, and 4, respectively. The vertex can be found using the formula:

• The x-coordinate is \( \frac{-b}{2a} \).
• Substitute \( a = -1 \) and \( b = 6 \) to get \( x = \frac{-6}{2(-1)} = 3 \).

To find the y-coordinate, substitute \( x = 3 \) back into the function: \( f(3) = -(3)^2 + 6(3) + 4 = 13 \). Hence, the vertex is located at point \((3, 13)\). The vertex indicates the maximum point because \( a < 0 \), and the parabola opens downwards.

X-Intercept

The x-intercepts of a quadratic function are the points where the graph crosses the x-axis. These are the solutions to the equation \( f(x) = 0 \). For our function, \( -x^2 + 6x + 4 = 0 \), you can apply the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
With the values \( a = -1 \), \( b = 6 \), \( c = 4 \), substitute them into the formula:

• Calculate \( b^2 - 4ac = 36 - 4(-1)(4) = 52 \).
• Find \( \sqrt{52} = 2\sqrt{13} \).
• The solutions are \( x = \frac{-6 \pm 2\sqrt{13}}{-2} = 3 \pm \sqrt{13} \).

Thus, the x-intercepts are \( x = 3 + \sqrt{13} \) and \( x = 3 - \sqrt{13} \). These points let us know where the parabola intersects the x-axis.

Y-Intercept

The y-intercept is the point where the graph crosses the y-axis. This is found by evaluating the quadratic function at \( x = 0 \). For our function \( f(x) = -x^2 + 6x + 4 \):

• Set \( x = 0 \) to get \( f(0) = -0^2 + 6(0) + 4 = 4 \).

Therefore, the y-intercept is \((0, 4)\). This value gives us a starting point for sketching and understanding the function's behavior along the y-axis.

Graph Sketching

Graph sketching visually represents the quadratic function, aiding in the understanding of its overall shape and key features. To successfully sketch the parabola of \( f(x) = -x^2 + 6x + 4 \):

• Begin by plotting the vertex at \((3, 13)\).
• Next, add the x-intercepts \((3 + \sqrt{13}, 0)\) and \((3 - \sqrt{13}, 0)\).
• Include the y-intercept at \((0, 4)\).

The parabola opens downwards since \( a \) is negative. Draw a symmetrical curve using the vertex as the axis of symmetry and the intercepts as key points.
Ensure to make the parabola pass through these critical points, with the vertex being the highest point, confirming its maximum value in this scenario. Sketching the graph helps visualize the quadratic function's behavior, emphasizing the direction it opens and the interactions with the axes.