Problem 18 - Intermediate Algebra

Question

Graph each of the following linear and quadratic functions. $$f(x)=2 x^{2}+12 x+14$$

Step-by-Step Solution

Answer

Graph the parabola with vertex at (-3, -4) and y-intercept at (0, 14); it opens upwards.

Step 1: Identify the Type of Equation

The given function is a quadratic function because it includes an \(x^2\) term. Quadratic functions are graphed as parabolas.

Step 2: Determine the Vertex

For a quadratic function in the form \(ax^2 + bx + c\), the vertex can be found using the formula \(x = -\frac{b}{2a}\). Here, \(a = 2\) and \(b = 12\), so \(x = -\frac{12}{2 \times 2} = -3\). Substitute \(x = -3\) back into the function to find the y-coordinate: \(f(-3) = 2(-3)^2 + 12(-3) + 14 = -4\). So the vertex is \((-3, -4)\).

Step 3: Determine the Axis of Symmetry

The axis of symmetry for the graph of a quadratic function is a vertical line through the vertex. For this function, the axis of symmetry is \(x = -3\).

Step 4: Find the Y-Intercept

The y-intercept of a function is the point where \(x = 0\). Substitute \(x = 0\) into the function: \(f(0) = 2(0)^2 + 12(0) + 14 = 14\). Thus, the y-intercept is \((0, 14)\).

Step 5: Identify Additional Points

Select values for \(x\) around the vertex to find additional points for the graph. For example, calculate \(f(-4)\) and \(f(-2)\). \(f(-4) = 2(-4)^2 + 12(-4) + 14 = 6\) and \(f(-2) = 2(-2)^2 + 12(-2) + 14 = 6\). These give points \((-4, 6)\) and \((-2, 6)\).

Step 6: Sketch the Graph

Plot the vertex \((-3, -4)\), the y-intercept \((0, 14)\), and additional points \((-4, 6)\) and \((-2, 6)\) on the graph. The parabola opens upwards since the coefficient of \(x^2\) is positive. Draw a symmetrical parabola through these points and along the axis of symmetry \(x = -3\).

Key Concepts

Parabolas

Parabolas are the U-shaped graphs of quadratic functions, recognized by their distinct shape. The term "parabola" specifically refers to the graph itself. Whenever you encounter a function in the form of \( ax^2 + bx + c \), you can anticipate a parabolic shape. Quadratic functions can open upwards or downwards. The direction depends on the coefficient \( a \):

• If \( a > 0 \), the parabola opens upwards, resembling a smile.
• If \( a < 0 \), it opens downwards, resembling a frown.

In our given function, \( f(x) = 2x^2 + 12x + 14 \), the coefficient \( a = 2 \), which is positive, thus confirming that the parabola opens upwards.

Vertex of a Parabola

The vertex of a parabola is its highest or lowest point, depending on its orientation. This characteristic makes the vertex an important feature as it represents either the maximum or minimum value of the quadratic function. Finding the vertex helps in understanding the overall shape and position of the parabola. The vertex can be calculated using the formula \( x = -\frac{b}{2a} \), which gives the x-coordinate of the vertex. Once you have this, substitute this x-value into the original function to find the corresponding y-coordinate. For our function \( f(x) = 2x^2 + 12x + 14 \):

• Here, \( a = 2 \) and \( b = 12 \), hence \( x = -\frac{12}{2 \times 2} = -3 \).
• To find the y-coordinate, substitute \( x = -3 \) back into the function: \( f(-3) = 2(-3)^2 + 12(-3) + 14 = -4 \).

Therefore, the vertex of the parabola is \((-3, -4)\), which serves as a pivotal reference point for graphing.

Axis of Symmetry

The axis of symmetry is a vertical line that divides the parabola into two mirrored halves. It is crucial for graphing because it indicates where the parabola is symmetric. In simple terms, if you fold the parabola along this line, both halves would align perfectly.To determine the axis of symmetry for a quadratic function, you can use the same x-value from the vertex formula, \( x = -\frac{b}{2a} \). This results in a vertical line described by a constant x-value. For our example, the axis of symmetry is:

• Through the vertex calculated earlier as \( x = -3 \), therefore, it is \( x = -3 \).

When graphing, draw this line along \( x = -3 \) to help position points symmetrically on either side of the axis.

Graphing Quadratic Functions

Graphing a quadratic function involves plotting its characteristic U-shape on a coordinate plane based on several key points and structural features. Start with these important components:

• The vertex, which we previously calculated as \((-3, -4)\), is your first and primary guide point.
• The axis of symmetry, a vital component in ensuring the parabola maintains its balanced shape, is drawn as a vertical line at \( x = -3 \).
• The y-intercept is the point where the parabola crosses the y-axis. It occurs when \( x = 0 \). For \( f(x) = 2x^2 + 12x + 14 \), the y-intercept is \((0, 14)\).
• Additional points enhance the accuracy of the graph. We calculated points \((-4, 6)\) and \((-2, 6)\) to aid symmetry.

Once these points are plotted, sketch the smooth curve of the parabola, ensuring it opens upwards due to the positive \( a \) value. The parabola should pass through all plotted points and reflect the symmetry across the axis.