Problem 11 - 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. $$f(x)=2 x^{2}+6 x$$

Step-by-Step Solution

Answer

The quadratic function is already in standard form. The vertex is \((-\frac{3}{2}, -\frac{9}{2})\), with \(x\)-intercepts \((0, 0)\), \((-3, 0)\), and a \(y\)-intercept \((0, 0)\).

Step 1: Express in Standard Form

The standard form of a quadratic function is \( ax^2 + bx + c \). The given function \( f(x) = 2x^2 + 6x \) already looks like it's in standard form where \( a = 2 \), \( b = 6 \), and \( c = 0 \).

Step 2: Find the Vertex

The vertex of a quadratic function in standard form \( ax^2 + bx + c \) is found using \( x = \frac{-b}{2a} \). For \( f(x) = 2x^2 + 6x \), \( a = 2 \) and \( b = 6 \). Therefore, \( x = \frac{-6}{2 \times 2} = -\frac{3}{2} \). Substitute \( x = -\frac{3}{2} \) back into the function to find \( y \): \( y = 2(-\frac{3}{2})^2 + 6(-\frac{3}{2}) = 2\times\frac{9}{4} - 9 = \frac{9}{2} - 9 = -\frac{9}{2} \). Thus, the vertex is \((-\frac{3}{2}, -\frac{9}{2})\).

Step 3: Find the x-intercepts

To find the \( x \)-intercepts, set \( f(x) = 0 \). So, \( 2x^2 + 6x = 0 \). Factor the equation: \( 2x(x + 3) = 0 \). Set each factor to zero: \( 2x = 0 \) or \( x + 3 = 0 \). Solving these gives \( x = 0 \) or \( x = -3 \). Therefore, the \( x \)-intercepts are \( (0, 0) \) and \( (-3, 0) \).

Step 4: Find the y-intercept

The \( y \)-intercept is found by setting \( x = 0 \) in the function. Substituting gives \( y = 2(0)^2 + 6(0) = 0 \). Thus, the \( y \)-intercept is \( (0, 0) \).

Step 5: Sketch the Graph

The function \( f(x) = 2x^2 + 6x \) is a parabola opening upwards because \( a = 2 \) is positive.Using the vertex \((-\frac{3}{2}, -\frac{9}{2})\), the \( x \)-intercepts \((0, 0)\) and \((-3, 0)\), and the \( y \)-intercept \((0, 0)\), sketch a symmetric parabola around the vertex, indicating these points on the graph.

Key Concepts

Standard Form of a Quadratic

A quadratic function can beautifully illustrate the shape of a parabola on a graph. It often appears in the standard form, expressed as \( ax^2 + bx + c \), where:

• \( a \) is the coefficient that determines the degree of the parabola’s width and direction.
• \( b \) influences the symmetry and linear component.
• \( c \) is the constant term that moves the parabola up or down on the graph.

For the function \( f(x) = 2x^2 + 6x \), the standard form is clear. Here, \( a = 2 \), \( b = 6 \), and \( c = 0 \). This standard appearance tells us important information about the function and prepares us to find other critical points like the vertex and intercepts.

Vertex of a Parabola

The vertex of a parabola is a key point. It is the highest or lowest point of the parabola depending on the direction it opens.To find the vertex for a quadratic in standard form, you use the formula \( x = \frac{-b}{2a} \). For our function \( f(x) = 2x^2 + 6x \), with \( a = 2 \) and \( b = 6 \), plug in the values:\[x = \frac{-6}{2 \times 2} = -\frac{3}{2}\]Next, substitute \( x = -\frac{3}{2} \) back into the function to get the \( y \)-value:\[y = 2(-\frac{3}{2})^2 + 6(-\frac{3}{2}) = -\frac{9}{2}\]Thus, the vertex is located at \((-\frac{3}{2}, -\frac{9}{2})\). This point is especially crucial for sketching because the parabola will have its ‘turnaround’ point here.

X-intercepts

X-intercepts are the points where the graph of the function crosses the x-axis. To find these for any function, set \( f(x) = 0 \) and solve for \( x \).With our example, \( f(x) = 2x^2 + 6x \), setting the equation to zero gives:\[2x^2 + 6x = 0\]Factoring out the common term, we have:\[2x(x + 3) = 0\]Setting each factor to zero:

• \( 2x = 0 \) gives \( x = 0 \)
• \( x + 3 = 0 \) gives \( x = -3 \)

Thus, the x-intercepts are \((0, 0)\) and \((-3, 0)\). These points are where the parabola touches or crosses the x-axis, playing a role in shaping the graph.

Y-intercept

The y-intercept is another important point. It’s where the parabola crosses the y-axis. Finding the y-intercept involves setting \( x = 0 \) in the function.For \( f(x) = 2x^2 + 6x \), substitute \( x = 0 \):\[y = 2(0)^2 + 6(0) = 0\]This means that the y-intercept is at \((0, 0)\). Interestingly, in this case, the y-intercept and one of the x-intercepts are the same point, which makes interpreting the graph a bit simpler. The y-intercept gives you an idea of where the parabola begins as it moves along the x-axis.