Problem 16 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 13-22: (a) Use the quadratic formula to find the zeros of \(f\). (b) Find the maximum or minimum value of \(f(x)\). (c) Sketch the graph of \(f\). $$ f(x)=6 x^{2}+7 x-24 $$

Step-by-Step Solution

Answer

Zeros are \(x = \frac{3}{2}\) and \(x = -\frac{8}{3}\). Minimum value of \(f(x)\) is \(-\frac{3750}{144}\). Graph is a parabola opening upwards.

Step 1: Identify the Coefficients

First, identify the coefficients in the quadratic function \(f(x) = ax^2 + bx + c\). Here, \(a = 6\), \(b = 7\), and \(c = -24\).

Step 2: Apply the Quadratic Formula

To find the zeros of \(f(x)\), use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute \(a = 6\), \(b = 7\), and \(c = -24\) into the formula.

Step 3: Calculate the Discriminant

Calculate the discriminant \(b^2 - 4ac\). \[b^2 - 4ac = 7^2 - 4(6)(-24) = 49 + 576 = 625\]

Step 4: Solve for the Zeros

With the discriminant calculated, solve for the zeros.\[x = \frac{-7 \pm \sqrt{625}}{12} = \frac{-7 \pm 25}{12}\]This gives two solutions for \(x\):\[x_1 = \frac{18}{12} = \frac{3}{2}, \quad x_2 = \frac{-32}{12} = -\frac{8}{3}\]

Step 5: Find the Vertex for Maximum/Minimum Value

The vertex form will give the maximum or minimum of the function, which can be found using \(x=-\frac{b}{2a}\).\[x = -\frac{7}{2 \times 6} = -\frac{7}{12}\]

Step 6: Calculate Maximum/Minimum Value

Substitute \(x = -\frac{7}{12}\) back into \(f(x)\).\[f\left(-\frac{7}{12}\right) = 6\left(-\frac{7}{12}\right)^2 + 7\left(-\frac{7}{12}\right) - 24\]Simplify this to find the function value.

Step 7: Simplify the Function Value

After substituting, calculate step-by-step:\[6\left(-\frac{7}{12}\right)^2 = 6 \times \frac{49}{144} = \frac{294}{144}\]\[7\left(-\frac{7}{12}\right) = -\frac{49}{12}\]Thus,\[ f\left(-\frac{7}{12}\right) = \frac{294}{144} - \frac{588}{144} - \frac{3456}{144} = -\frac{3750}{144} \]

Step 8: Sketch the Graph

Use the zeros \(x_1 = \frac{3}{2}\) and \(x_2 = -\frac{8}{3}\), along with the vertex \((-\frac{7}{12}, -\frac{3750}{144})\), to sketch the parabola opening upwards (since \(a > 0\)). Note that the graph is a U-shape with these intersections and a minimum point.

Key Concepts

Quadratic Formula

The quadratic formula is an essential tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). This handy formula allows us to find the zeros of a quadratic function, that is, the values of \( x \) where the function equals zero.
Use the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Plug in the coefficients \( a = 6 \), \( b = 7 \), and \( c = -24 \) to find the zeros of \( f(x) = 6x^2 + 7x - 24 \). You can calculate each part separately:

• Calculate \( -b \), which equals \(-7\).
• Find the discriminant \( \sqrt{b^2 - 4ac} \), which we'll cover below.

Finally, divide by \( 2a \), which in this case equals \( 12 \). Solve for each case—adding and subtracting the square root—to find both solutions.

Discriminant

The discriminant is one part of the quadratic formula, specifically \( b^2 - 4ac \). This component helps us understand more about the nature of the quadratic function's roots without solving it completely.
The discriminant can tell us:

• If \( b^2 - 4ac > 0 \), there are two real and distinct solutions (or zeros).
• If \( b^2 - 4ac = 0 \), there's exactly one real solution—meaning the parabola touches the x-axis at one point.
• If \( b^2 - 4ac < 0 \), there are no real solutions; the parabola does not intersect the x-axis.

In our example, \( b^2 - 4ac = 625 \), indicating the presence of two distinct real roots. This makes sketching the function easier since it assures us of two intersections with the x-axis.

Vertex Form

Although the quadratic formula is excellent for finding zeros, finding the vertex can help with determining the maximum or minimum value of a quadratic function.
The vertex form of a quadratic function is another expression often given by \( f(x) = a(x - h)^2 + k \), where \((h, k)\) represents the vertex of the parabola. To find the vertex from standard form \( ax^2 + bx + c \), we use:

• \( x = -\frac{b}{2a} \), which gives us the x-coordinate of the vertex.

Substitute this back into \( f(x) \) to find the corresponding y-value, \( k \). In our specific function, substituting \( x = -\frac{7}{12} \) back into \( f(x) \), gives us the minimum value because the parabola opens upwards (\( a > 0 \)). This vertex guides us in sketching the parabola correctly.

Zeros of a Function

The zeros of a function are the x-values where the function itself equals zero. These are critical points on the graph where it intersects the x-axis and are also known as the solutions of the equation.
Using the quadratic formula, for \( f(x) = 6x^2 + 7x - 24 \), we found:

• \( x_1 = \frac{3}{2} \)
• \( x_2 = -\frac{8}{3} \)

These zeros provide essential information for understanding the overall shape and position of the parabola. The presence of two zeros indicates the function touches the x-axis at two points.
Zeros are especially important when graphing, as they demonstrate where the graph crosses or touches the x-axis, helping to visualize the function's behavior across different intervals.