5) The graph of the quadratic function f(x) = (x - 1)(x + 5) intersects the y-axis (0, c). The vertex of th function is (-2,-6.75). The equation has two x values where f(x) = -6 and the first x value is at x = −3 a) Find the value of c. (0) = 2/1 (0-1) (075) (1)(5) (M²) -5 b) Write down the equation for the axis of symmetry of the graph. 2 P(x) = 3/1 (x-7)(x + 5) 3/ = 3/4x²+5x-x-5 c) Write down the x-intercepts of t graph. 6 * x-1=0 X=1 d) Draw the graph of y = f(x) for -7 ≤ x ≤ 3 and -7 ≤ y ≤ 12 (4₂) x²+x-5 X+5=0 x= -√ 3.75 e) Use the symmetry of the graph to show that the second solution is x = -1.
5) The graph of the quadratic function f(x) = (x - 1)(x + 5) intersects the y-axis (0, c). The vertex of th function is (-2,-6.75). The equation has two x values where f(x) = -6 and the first x value is at x = −3 a) Find the value of c. (0) = 2/1 (0-1) (075) (1)(5) (M²) -5 b) Write down the equation for the axis of symmetry of the graph. 2 P(x) = 3/1 (x-7)(x + 5) 3/ = 3/4x²+5x-x-5 c) Write down the x-intercepts of t graph. 6 * x-1=0 X=1 d) Draw the graph of y = f(x) for -7 ≤ x ≤ 3 and -7 ≤ y ≤ 12 (4₂) x²+x-5 X+5=0 x= -√ 3.75 e) Use the symmetry of the graph to show that the second solution is x = -1.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Can someone please help with B, D, and E please? Thank you.
![Sure! Here's a transcription suitable for an educational website:
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**Question 5: Analysis of a Quadratic Function**
The graph of the quadratic function \( f(x) = \frac{3}{4}(x-1)(x+5) \) intersects the y-axis at the point \((0, c)\). The vertex of this function is \((-2, -6.75)\). The equation has two \(x\) values where \(f(x) = -6\), and the first \(x\) value is at \(x = -3\).
**Tasks:**
**a) Find the value of \(c\).**
Solution:
To find \(c\), substitute \(x = 0\) into the function:
\[ f(0) = \frac{3}{4}(0-1)(0+5) = \frac{3}{4}(-1)(5) = -\frac{15}{4} = -3.75 \]
Thus, \(c = -3.75\).
**b) Write down the equation for the axis of symmetry of the graph.**
Solution:
The axis of symmetry can be found using the average of the x-intercepts:
\[ \text{Equation of axis of symmetry: } x = -\frac{1}{2}(1 + (-5)) = -2 \]
**c) Write down the x-intercepts of the graph.**
Solution:
The x-intercepts are found by setting \( f(x) = 0 \):
\[
\begin{align*}
\frac{3}{4}(x-1)(x+5) &= 0 \\
x - 1 &= 0 \quad \Rightarrow \quad x = 1 \\
x + 5 &= 0 \quad \Rightarrow \quad x = -5
\end{align*}
\]
Thus, the x-intercepts are \( x = 1 \) and \( x = -5 \).
**d) Draw the graph of \( y = f(x) \) for \(-7 \leq x \leq 3\) and \(-7 \leq y \leq 12\).**
(Graph illustration)
Description:
- The graph is a parabola opening downwards, intersecting the y-axis at \( (0, -3.75) \).
- The x-intercepts are at \( (1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8beeba18-127d-4f79-b572-133114d91afd%2F76b7201d-84de-4a96-815b-b2366b75c9dd%2Fucx4xl9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Sure! Here's a transcription suitable for an educational website:
---
**Question 5: Analysis of a Quadratic Function**
The graph of the quadratic function \( f(x) = \frac{3}{4}(x-1)(x+5) \) intersects the y-axis at the point \((0, c)\). The vertex of this function is \((-2, -6.75)\). The equation has two \(x\) values where \(f(x) = -6\), and the first \(x\) value is at \(x = -3\).
**Tasks:**
**a) Find the value of \(c\).**
Solution:
To find \(c\), substitute \(x = 0\) into the function:
\[ f(0) = \frac{3}{4}(0-1)(0+5) = \frac{3}{4}(-1)(5) = -\frac{15}{4} = -3.75 \]
Thus, \(c = -3.75\).
**b) Write down the equation for the axis of symmetry of the graph.**
Solution:
The axis of symmetry can be found using the average of the x-intercepts:
\[ \text{Equation of axis of symmetry: } x = -\frac{1}{2}(1 + (-5)) = -2 \]
**c) Write down the x-intercepts of the graph.**
Solution:
The x-intercepts are found by setting \( f(x) = 0 \):
\[
\begin{align*}
\frac{3}{4}(x-1)(x+5) &= 0 \\
x - 1 &= 0 \quad \Rightarrow \quad x = 1 \\
x + 5 &= 0 \quad \Rightarrow \quad x = -5
\end{align*}
\]
Thus, the x-intercepts are \( x = 1 \) and \( x = -5 \).
**d) Draw the graph of \( y = f(x) \) for \(-7 \leq x \leq 3\) and \(-7 \leq y \leq 12\).**
(Graph illustration)
Description:
- The graph is a parabola opening downwards, intersecting the y-axis at \( (0, -3.75) \).
- The x-intercepts are at \( (1
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