Problem 38 - College Algebra with Corequisite Support

Question

For the following exercises, sketch a graph of the quadratic function and give the vertex, axis of symmetry, and intercepts. $$ f(x)=-2 x^{2}+5 x-8 $$

Step-by-Step Solution

Answer

Vertex: \(\left(\frac{5}{4}, -\frac{81}{8}\right)\); Axis: \(x=\frac{5}{4}\); Intercepts: y-intercept \((0, -8)\), no real x-intercepts.

Step 1: Identify the Coefficients

The quadratic function given is \( f(x) = -2x^2 + 5x - 8 \). The coefficients are \( a = -2 \), \( b = 5 \), and \( c = -8 \).

Step 2: Find the Vertex

The vertex of a quadratic function \( ax^2 + bx + c \) is given by \( x = \frac{-b}{2a} \). Substituting the values, \( x = \frac{-5}{2(-2)} = \frac{5}{4} \). To find \( y \), substitute \( \frac{5}{4} \) into the function: \( f\left(\frac{5}{4}\right) = -2\left(\frac{5}{4}\right)^2 + 5\left(\frac{5}{4}\right) - 8 \). Calculating gives \( f\left(\frac{5}{4}\right) = -\frac{81}{8} \). Hence, the vertex is \( \left( \frac{5}{4}, -\frac{81}{8} \right) \).

Step 3: Determine the Axis of Symmetry

The axis of symmetry is a vertical line that passes through the vertex. Thus, the axis of symmetry is \( x = \frac{5}{4} \).

Step 4: Find the Y-intercept

The y-intercept of a quadratic function is the point where \( x = 0 \). Substitute \( x = 0 \) in \( f(x) = -2(0)^2 + 5(0) - 8 \) which gives \( f(0) = -8 \). Therefore, the y-intercept is \((0, -8)\).

Step 5: Find the X-intercepts

The x-intercepts occur where \( f(x) = 0 \). So, solve \( -2x^2 + 5x - 8 = 0 \) using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substituting the coefficients gives \( x = \frac{-5 \pm \sqrt{5^2 - 4(-2)(-8)}}{2(-2)} \). Calculate \( x = \frac{-5 \pm \sqrt{25 - 64}}{-4} = \frac{-5 \pm \sqrt{-39}}{-4} \). As the discriminant is negative, there are no real x-intercepts.

Step 6: Sketch the Graph

Plot the vertex \( \left( \frac{5}{4}, -\frac{81}{8} \right) \), the axis of symmetry \( x = \frac{5}{4} \), and the y-intercept \( (0, -8) \) on a coordinate plane. Since there are no real x-intercepts, the parabola does not cross the x-axis. Draw the parabola opening downward because \( a = -2 \), which is less than zero, indicating it is upside-down.

Key Concepts

Vertex of a Parabola

In quadratic functions, the vertex is a fundamental point. It serves as the "turning" or "peak" point of the parabola. For a function in standard form, \( f(x) = ax^2 + bx + c \), the vertex can be found using the formula \( x = \frac{-b}{2a} \).
To find the y-coordinate, you substitute the x-value back into the function. What results is the point \( (x, y) \), which represents the vertex. This particular point indicates where the graph changes direction.
In the given function \( f(x) = -2x^2 + 5x - 8 \), solving for \( x \) using the equation \( x = \frac{-b}{2a} \) results in \( x = \frac{5}{4} \). Next, substitute this \( x \)-value back into the function to find \( y \), yielding \( f\left(\frac{5}{4}\right) = -\frac{81}{8} \). Thus, the vertex is \( \left( \frac{5}{4}, -\frac{81}{8} \right) \).

Axis of Symmetry

The axis of symmetry of a parabola is an essential feature that reflects its symmetrical nature. This axis is a vertical line that divides the parabola into two mirror-image halves. It always passes through the vertex of the parabola.
The equation of the axis of symmetry is derived from the vertex's x-coordinate. For our function, the axis is given by \( x = \frac{5}{4} \).
Knowing the axis of symmetry helps in mirroring points across to graph the parabola effectively. In simpler terms, if you fold the parabola along this axis, both sides match perfectly. This characteristic of symmetry is quite useful when sketching or analyzing quadratic graphs.

Graphing Quadratics

Graphing a quadratic function involves plotting specific points and understanding how the parabola behaves. Here's a step-by-step:

• Begin by identifying the vertex. This gives a central reference point for the graph.
• Understand that the parabola opens upwards if \( a > 0 \) and downwards if \( a < 0 \). Our quadratic has \( a = -2 \), so the parabola opens downwards.
• Plot the vertex and y-intercept on the coordinate plane.
• Use the axis of symmetry to ensure the parabola is balanced."

Graphing is invaluable for visualizing how a function behaves, which can enhance understanding of its roots, intercepts, and general shape.

Intercepts of a Quadratic

Intercepts are the points where a parabola crosses the axes. The y-intercept is the point where the graph crosses the y-axis. It's found by setting \( x = 0 \) in the function.
For the function \( f(x) = -2x^2 + 5x - 8 \), substituting \( x = 0 \) gives \( y = -8 \), indicating a y-intercept at \( (0, -8) \).
For x-intercepts, you look for points where \( f(x) = 0 \). This involves solving the equation \( ax^2 + bx + c = 0 \). Using the quadratic formula, we find that there are no real x-intercepts if the discriminant \( (b^2 - 4ac) \) is negative. For our function, since the discriminant is \( 25 - 64 = -39 \), there are no real roots. This indicates the parabola doesn't cross the x-axis, consistent with the graph we've plotted.