Problem 8 - Algebra and Trigonometry

Question

\(5-18=\) 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. $$ f(x)=-x^{2}+10 x $$

Step-by-Step Solution

Answer

Vertex: (5, 25); x-intercepts: (0, 0), (10, 0); y-intercept: (0, 0).

Step 1: Express in Standard Form

The standard form of a quadratic function is \( f(x) = ax^2 + bx + c \). The given function is already in standard form: \( f(x) = -x^2 + 10x + 0 \) where \( a = -1 \), \( b = 10 \), and \( c = 0 \).

Step 2: Find the Vertex

The vertex form of a quadratic is \( h = \frac{-b}{2a} \) for the \( x \)-coordinate, and plugging \( h \) back into the function to find \( k \). So here, \( h = \frac{-10}{2(-1)} = 5 \). Substituting \( h \) into the function, \( f(5) = -(5)^2 + 10(5) = 25 \). Thus, the vertex is \((5, 25)\).

Step 3: Find the X-Intercepts

The \( x \)-intercepts occur where \( f(x) = 0 \). Solve \( -x^2 + 10x = 0 \) which factors to \( x(-x + 10) = 0 \). Thus, \( x = 0 \) or \( x = 10 \). The \( x \)-intercepts are at \( x = 0 \) and \( x = 10 \).

Step 4: Find the Y-Intercept

The \( y \)-intercept is the value of \( f(x) \) when \( x = 0 \). Here, \( f(0) = 0 \), so the \( y \)-intercept is \( 0 \).

Step 5: Sketch the Graph

To sketch the graph, plot the vertex at \((5, 25)\) and the intercepts: the \( x \)-intercepts at \( (0, 0) \) and \( (10, 0) \), and the \( y \)-intercept at \((0,0)\). Connect these with a parabolic curve opening downwards (since \( a = -1 \)).

Key Concepts

Vertex of a Quadratic Function

The vertex of a quadratic function provides the maximum or minimum point on its graph. Understanding how to find the vertex can give insight into the behavior and characteristics of the quadratic function. Quadratic functions have the form \( f(x) = ax^2 + bx + c \) when expressed in standard form. Finding the vertex involves determining specific coordinates, often denoted as \((h, k)\).

To locate the vertex:

• Calculate \( h \), the x-coordinate, using the formula \( h = \frac{-b}{2a} \).
• Substitute \( h \) back into the function to find \( k \), the y-coordinate, using \( f(h) = k \).

In our exercise, for the function \( f(x) = -x^2 + 10x \), we determined \( h = 5 \) and \( k = 25 \) by substituting this x-value back into the equation. Therefore, the vertex is \((5, 25)\). This specific point helps us understand that the parabola reaches its peak at this point before descending, given that the parabola opens downward due to a negative \( a \).

X-Intercepts

X-intercepts are the points where the graph of a quadratic function crosses the x-axis. At these points, the value of the function is zero, i.e., \( f(x) = 0 \). To find the x-intercepts, you solve the function equation for zero.

In the quadratic equation from our exercise, \( f(x) = -x^2 + 10x \), we set the function equal to zero: \(-x^2 + 10x = 0\). By factoring, we find \( x(x - 10) = 0 \), leading to solutions \( x = 0 \) and \( x = 10 \).

• The first intercept is at \((0, 0)\).
• The second is at \((10, 0)\).

The x-intercepts reveal where the function meets the x-axis and how it behaves across this axis. They mark points of entry and exit of the parabola on the x-plane.

Y-Intercepts

The y-intercept of a quadratic function is where the function's graph intersects the y-axis. This point occurs when all terms involving \( x \) equate to zero, essentially when \( x = 0 \).

For our function, \( f(x) = -x^2 + 10x \), let's substitute \( x = 0 \) into the equation, resulting in \( f(0) = 0 \). Hence, the y-intercept is at \((0, 0)\), which coincidentally aligns with one of the x-intercepts.

• The y-intercept is simply the value of the function at \( x = 0 \).
• It serves as an anchor point on the y-axis.

In describing the curve's shape, understanding intercepts assists in predicting how the function transitions from quadrant to quadrant on the graph.

Standard Form of a Quadratic Function

The standard form is a way of expressing quadratic functions using the template \( f(x) = ax^2 + bx + c \). This structure highlights the quadratic, linear, and constant terms.

To express a function in standard form, identify coefficients:

• \( a \) is the coefficient of \( x^2 \).
• \( b \) provides the linear component with \( x \).
• \( c \) is the constant term.

The provided quadratic, \( f(x) = -x^2 + 10x \), is already in standard form with \( a = -1 \), \( b = 10 \), and \( c = 0 \). This layout aids in locating the vertex, intercepts, and sketching the parabola's general shape. Recognizing a quadratic in its standard form simplifies further calculations and visualizations of the function's behavior.