Question
*Problem Statement*
In the diagram, the graphs of $f(x) = -x^2 + 5x + 6$ and $g(x)$ intersect the graph of $f(x)$ at $B$ and $C$. $g(x)$ is perpendicular to the x-axis with points $P$ and $Q$ on $f(x)$ and $g(x)$ respectively. $M$ is the TP of $f(x)$.
*
1 calculate:
Co ordinates of M
Maximum length of PQ between bend s
Answers
GPT-4o mini
To solve the problem, we need to first handle the quadratic function \( f(x) = -x^2 + 5x + 6 \).
### Step 1: Finding the coordinates of the Maximum Point (M) of \( f(x) \)
The function \( f(x) \) is a downward-opening parabola (since the coefficient of \( x^2 \) is negative). The x-coordinate of the vertex (maximum point) of a quadratic function \( ax^2 + bx + c \) is given by:
\[
x = -\frac{b}{2a}
\]
Here, \( a = -1 \) and \( b = 5 \):
\[
x = -\frac{5}{2 \times -1} = \frac{5}{2} = 2.5
\]
Now, we substitute \( x = 2.5 \) back into \( f(x) \) to find the y-coordinate of point M:
\[
f(2.5) = - (2.5)^2 + 5(2.5) + 6
\]
Calculating \( f(2.5) \):
\[
f(2.5) = -6.25 + 12.5 + 6 = -6.25 + 18.5 = 12.25
\]
Thus, the coordinates of M (the vertex) are:
\[
\text{M} = \left(2.5, 12.25\right)
\]
### Step 2: Finding the Maximum Length of Segment PQ between the bends
The function \( f(x) \) can be rewritten in its vertex form, which we already obtained, or we could determine the points where the tangent to the curve is horizontal. The bends of the function are important for the segment \( PQ \).
The derivative \( f'(x) \) gives us the slope of the tangent line:
\[
f'(x) = -2x + 5 = 0 \implies x = \frac{5}{2} = 2.5
\]
Since we know \( M = (2.5, 12.25) \), and \( g(x) \) intersects \( f(x) \) perpendicularly at points \( P \) and \( Q \) (line g being vertical).
To find points \( P \) and \( Q \), we need the x-coordinates on both sides of \( M \) where \( f(x) \) is still decreasing (with respect to \( M \)) until it crosses the x-axis.
### Step 3: Finding the x-intercepts of \( f(x) \)
We set \( f(x) = 0 \):
\[
-x^2 + 5x + 6 = 0 \implies x^2 - 5x - 6 = 0
\]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
x = \frac{5 \pm \sqrt{25 + 24}}{2} = \frac{5 \pm 7}{2}
\]
Calculating the roots:
\[
x_1 = \frac{12}{2} = 6 \quad \text{and} \quad x_2 = \frac{-2}{2} = -1
\]
So, the x-intercepts are at \( x = 6 \) and \( x = -1 \).
### Step 4: Maximum Length of PQ
The maximum segment \( PQ \) occurs between these intercepts. Since \( g(x) \) is perpendicular to the x-axis, \( P \) could be anywhere along the peaks of \( f(x) \) between the roots. Thus, \( P \) could be at the max point \( M(2.5, 12.25)\). The vertical distance can be calculated using \( g(x) \).
The maximum length \( |PQ| \) can be calculated from the maximum height at \( M \):
From \( M \) to points at \( f(-1) \) and \( f(6) \):
At \( x = -1 \):
\[
f(-1) = -(-1)^2 + 5(-1) + 6 = -1 - 5 + 6 = 0
\]
At \( x = 6 \):
\[
f(6) = -(6^2) + 5(6) + 6 = -36 + 30 + 6 = 0
\]
Thus, \( PQ \) lies vertically from maximum height \( f(-1) = 0 \) and \( f(6) = 0 \).
Maximum length \( PQ = M_{\text{height}} = 12.25 \).
### Conclusion
1. Coordinates of M: \(\mathbf{(2.5, 12.25)}\).
2. Maximum length of PQ: \(\mathbf{12.25}\).
### Step 1: Finding the coordinates of the Maximum Point (M) of \( f(x) \)
The function \( f(x) \) is a downward-opening parabola (since the coefficient of \( x^2 \) is negative). The x-coordinate of the vertex (maximum point) of a quadratic function \( ax^2 + bx + c \) is given by:
\[
x = -\frac{b}{2a}
\]
Here, \( a = -1 \) and \( b = 5 \):
\[
x = -\frac{5}{2 \times -1} = \frac{5}{2} = 2.5
\]
Now, we substitute \( x = 2.5 \) back into \( f(x) \) to find the y-coordinate of point M:
\[
f(2.5) = - (2.5)^2 + 5(2.5) + 6
\]
Calculating \( f(2.5) \):
\[
f(2.5) = -6.25 + 12.5 + 6 = -6.25 + 18.5 = 12.25
\]
Thus, the coordinates of M (the vertex) are:
\[
\text{M} = \left(2.5, 12.25\right)
\]
### Step 2: Finding the Maximum Length of Segment PQ between the bends
The function \( f(x) \) can be rewritten in its vertex form, which we already obtained, or we could determine the points where the tangent to the curve is horizontal. The bends of the function are important for the segment \( PQ \).
The derivative \( f'(x) \) gives us the slope of the tangent line:
\[
f'(x) = -2x + 5 = 0 \implies x = \frac{5}{2} = 2.5
\]
Since we know \( M = (2.5, 12.25) \), and \( g(x) \) intersects \( f(x) \) perpendicularly at points \( P \) and \( Q \) (line g being vertical).
To find points \( P \) and \( Q \), we need the x-coordinates on both sides of \( M \) where \( f(x) \) is still decreasing (with respect to \( M \)) until it crosses the x-axis.
### Step 3: Finding the x-intercepts of \( f(x) \)
We set \( f(x) = 0 \):
\[
-x^2 + 5x + 6 = 0 \implies x^2 - 5x - 6 = 0
\]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
x = \frac{5 \pm \sqrt{25 + 24}}{2} = \frac{5 \pm 7}{2}
\]
Calculating the roots:
\[
x_1 = \frac{12}{2} = 6 \quad \text{and} \quad x_2 = \frac{-2}{2} = -1
\]
So, the x-intercepts are at \( x = 6 \) and \( x = -1 \).
### Step 4: Maximum Length of PQ
The maximum segment \( PQ \) occurs between these intercepts. Since \( g(x) \) is perpendicular to the x-axis, \( P \) could be anywhere along the peaks of \( f(x) \) between the roots. Thus, \( P \) could be at the max point \( M(2.5, 12.25)\). The vertical distance can be calculated using \( g(x) \).
The maximum length \( |PQ| \) can be calculated from the maximum height at \( M \):
From \( M \) to points at \( f(-1) \) and \( f(6) \):
At \( x = -1 \):
\[
f(-1) = -(-1)^2 + 5(-1) + 6 = -1 - 5 + 6 = 0
\]
At \( x = 6 \):
\[
f(6) = -(6^2) + 5(6) + 6 = -36 + 30 + 6 = 0
\]
Thus, \( PQ \) lies vertically from maximum height \( f(-1) = 0 \) and \( f(6) = 0 \).
Maximum length \( PQ = M_{\text{height}} = 12.25 \).
### Conclusion
1. Coordinates of M: \(\mathbf{(2.5, 12.25)}\).
2. Maximum length of PQ: \(\mathbf{12.25}\).