Problem 17 - 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. $$f(x)=2 x^{2}+4 x+3$$

Step-by-Step Solution

Answer

The vertex is (-1, 1); no x-intercepts; y-intercept is (0, 3).

Step 1: Write the quadratic in standard form

The given quadratic function is already in standard form, which is \( f(x) = ax^2 + bx + c \). For this function, \( a = 2 \), \( b = 4 \), and \( c = 3 \).

Step 2: Calculate the Vertex

The vertex of a quadratic function \( f(x) = ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \) for the x-coordinate. Here, \( b = 4 \) and \( a = 2 \), so the x-coordinate of the vertex is \(-\frac{4}{4} = -1\). Substitute \( x = -1 \) back into the function to find the y-coordinate: \( f(-1) = 2(-1)^2 + 4(-1) + 3 = 2 - 4 + 3 = 1 \). Thus, the vertex is \((-1, 1)\).

Step 3: Find the x-intercepts

To find the x-intercepts, set \( f(x) = 0 \) and solve for \( x \): \[ 2x^2 + 4x + 3 = 0 \] Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 2 \), \( b = 4 \), \( c = 3 \): \[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 2 \cdot 3}}{2 \cdot 2} = \frac{-4 \pm \sqrt{16 - 24}}{4} = \frac{-4 \pm \sqrt{-8}}{4} \] Since the discriminant is negative, there are no x-intercepts (the graph does not cross the x-axis).

Step 4: Find the y-intercept

The y-intercept of a function is found by evaluating \( f(x) \) at \( x = 0 \). Substitute \( x = 0 \): \[ f(0) = 2(0)^2 + 4(0) + 3 = 3 \] Thus, the y-intercept is \( (0, 3) \).

Step 5: Sketch the graph

To sketch the graph, plot the vertex \((-1, 1)\) and the y-intercept \((0, 3)\) on the coordinate plane. Draw a parabola opening upwards (since \( a = 2 > 0 \)), passing through these points. The parabola does not cross the x-axis but passes above it.

Key Concepts

Vertex Calculation

Understanding vertex calculation is crucial when dealing with quadratic functions in the form of \( f(x) = ax^2 + bx + c \). The vertex is a significant point since it represents the minimum or maximum of the parabola, depending on whether it opens upwards or downwards. To find the vertex, use the formula \( x = -\frac{b}{2a} \) to determine the x-coordinate.

In our function \( f(x) = 2x^2 + 4x + 3 \), \( a = 2 \) and \( b = 4 \). Substitute these values into the formula:

• \( x = -\frac{4}{2 \times 2} = -1 \)

Now, plug \( x = -1 \) back into the function to find the y-coordinate of the vertex:

• \( f(-1) = 2(-1)^2 + 4(-1) + 3 = 2 - 4 + 3 = 1 \)

Therefore, the vertex of the parabola is \((-1, 1)\). This point indicates both the lowest point on the graph because \( a > 0 \), reflecting the parabola's upward opening.

X-Intercept

X-intercepts are the points where the graph of the function crosses the x-axis. For those points, the function value is zero, so we need to solve \( f(x) = 0 \).

Given the quadratic function \( f(x) = 2x^2 + 4x + 3 \), set it to zero:

• \( 2x^2 + 4x + 3 = 0 \)

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve.

• \( a = 2 \), \( b = 4 \), \( c = 3 \)
• Calculate the discriminant: \( b^2 - 4ac = 16 - 24 = -8 \)

Since the discriminant is negative, \( \sqrt{-8} \) is complex, meaning there are no real solutions and thus no x-intercepts. The graph does not cross the x-axis, indicating that the entire parabola is located above the x-axis.

Y-Intercept

Calculating the y-intercept of a quadratic function is straightforward. It is the point where the graph intersects the y-axis, which occurs when \( x = 0 \). To find this point, substitute \( x = 0 \) into the quadratic function.

• For \( f(x) = 2x^2 + 4x + 3 \), substitute to get \( f(0) = 3 \)

Therefore, the y-intercept is the point \((0, 3)\). This point represents the height where the parabola initially intersects the y-axis, giving us a point of reference on the vertical axis when plotting the graph.

Graph Sketching

Sketching the graph of a quadratic function involves plotting its critical points and establishing the curve's shape. Here's how you can do it:First, plot the vertex and the y-intercept on the coordinate plane.

• Vertex: \((-1, 1)\)
• Y-intercept: \((0, 3)\)

The absence of x-intercepts tells us the parabola does not cross the x-axis.

Since the leading coefficient \( a = 2 \) is positive, the parabola opens upwards. This shapes the curve above the x-axis, with the vertex at its lowest point.

Draw a smooth curve starting from the vertex, passing through the y-intercept, and extending upwards symmetrically on both sides. Make sure the graph depicts a U-shape, typical of quadratic functions with positive \( a \). This visualization helps understand how the function behaves and confirms characteristics like vertex position and intercepts.