Problem 79 - Algebra and Trigonometry

Question

Based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. If \(f(x)=a x^{2}-2 x+5\) and \(a<0,\) in which quadrant is the vertex located? How many \(x\) -intercepts does the graph of \(f\) have?

Step-by-Step Solution

Answer

The vertex is in the second quadrant and the graph has two x-intercepts.

Step 1: Identify the vertex formula

The vertex of a quadratic function in the form of \(f(x) = ax^2 + bx + c\) can be found using the formula for the x-coordinate: \(x = \frac{-b}{2a}\). Here, \(a = a\), \(b = -2\), and \(c = 5\).

Step 2: Calculate the x-coordinate of the vertex

Substitute \(b = -2\) and \(a < 0\) into the formula \(x = \frac{-b}{2a}\): \[ x = \frac{-(-2)}{2a} = \frac{2}{2a} = \frac{1}{a} \]

Step 3: Determine the y-coordinate of the vertex

Substitute \(x = \frac{1}{a}\) back into the function \(f(x)\) to find the y-coordinate: \[ f \left( \frac{1}{a} \right) = a \left( \frac{1}{a} \right)^2 - 2 \left( \frac{1}{a} \right) + 5 = \frac{1}{a} - \frac{2}{a} + 5 = \frac{1 - 2}{a} + 5 = \frac{-1}{a} + 5 \] Since \(a < 0\), \( \frac{-1}{a} \) is positive. Thus, the y-coordinate is \(\frac{-1}{a} + 5 > 5\).

Step 4: Locate the vertex quadrant

Since the x-coordinate \(\frac{1}{a}\) is negative when \(a < 0\), and the y-coordinate \(\frac{-1}{a} + 5\) is positive, the vertex is located in the second quadrant.

Step 5: Find the discriminant for the x-intercepts

The number of x-intercepts can be determined by the discriminant \(\Delta = b^2 - 4ac\). Substitute \(a = a\), \(b = -2\), and \(c = 5\) into the discriminant formula: \[ \Delta = (-2)^2 - 4(a)(5) = 4 - 20a \] Since \(a < 0\), \(20a < 0\), thus \(4 - 20a > 4\). This means \(\Delta > 0\), indicating that there are two x-intercepts.

Key Concepts

Vertex Formula

To find the vertex of a quadratic function, we use the vertex formula for the x-coordinate: \[ x = \frac{-b}{2a} \] where the quadratic function is in the form of \[ f(x) = ax^2 + bx + c. \] This formula helps pinpoint the x-coordinate of the vertex.
For the given function \[ f(x) = ax^2 - 2x + 5 \] with \[ a < 0 \], we identify \- \[a\] = \[a\] and \[-b\] = \[-(-2)\].
By substituting \[-b\] and \[2a\] into the formula, we get \[ x = \frac{2}{2a} = \frac{1}{a}. \] The next step involves finding the corresponding y-coordinate. This requires substituting \[ x = \frac{1}{a} \] back into the original function to compute \[ f \left( \frac{1}{a} \right). \] After calculations, this gives \[ f \left( \frac{1}{a} \right) = \frac{-1}{a} + 5. \] Since \[ a < 0 \], the factor \[ \frac{-1}{a}} \] becomes positive, resulting in a y-coordinate greater than 5.

X-intercepts

The x-intercepts of a quadratic function are the points where the graph intersects the x-axis.
To find the x-intercepts, we solve for \[ f(x) = 0. \] This involves using the quadratic formula: \[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \] where the discriminant (\[\Delta\]) is \[ b^2 - 4ac. \]
In our specific function \[ f(x) = ax^2 - 2x + 5 \] with \[ a < 0 \], the parameters are \[ a = a, b = -2, c = 5. \] Computing the discriminant gives: \[ \Delta = (-2)^2 - 4(a)(5) = 4 - 20a \] Since \[ a < 0 \], the term \[ 20a \] will be negative. Thus, \[ 4 - 20a \] results in a value greater than 4, making \[ \Delta > 0 \].
This positive discriminant indicates that the quadratic function has two distinct x-intercepts.

Discriminant

The discriminant (\[\Delta\]) in a quadratic equation \[ ax^2 + bx + c = 0 \] provides vital information about the nature of its roots.
The discriminant is calculated as \[ \Delta = b^2 - 4ac. \] Depending on whether \[ \Delta \] is positive, zero, or negative, the quadratic equation will have:

• Two distinct real roots: If \[ \Delta > 0 \]
• One real root: If \[ \Delta = 0 \]
• No real roots: If \[ \Delta < 0 \]

For the quadratic function \[ f(x) = ax^2 - 2x + 5 \] with \[ a < 0 \], the parameters are \[ a = a, b = -2, c = 5. \] Plugging these into the discriminant formula gives: \[ \Delta = (-2)^2 - 4(a)(5) = 4 - 20a. \] Since \[ a < 0 \], the term \[ 20a \] is negative, resulting in \[ \Delta > 0. \] Consequently, the quadratic equation has two x-intercepts, meaning it crosses the x-axis at two points.