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Let's tackle each part of the question methodically.
### 5.1 Coordinates of [tex]\( E \)[/tex]
The turning point of a quadratic function in the form [tex]\( f(x) = a(x - h)^2 + k \)[/tex] is at the point [tex]\((h, k)\)[/tex]. Given:
[tex]\[ f(x) = -\left(x - \frac{7}{2}\right)^2 + \frac{81}{4} \][/tex]
The vertex, which is the turning point [tex]\( E \)[/tex], has the coordinates:
[tex]\[ E \left( \frac{7}{2}, \frac{81}{4} \right) \][/tex]
### 5.2 Average Gradient of [tex]\( f \)[/tex] between [tex]\( x = 1 \)[/tex] and [tex]\( x = 5 \)[/tex]
The average gradient of a function between two points [tex]\( x=a \)[/tex] and [tex]\( x=b \)[/tex] is calculated as:
[tex]\[ \text{Average Gradient} = \frac{f(b) - f(a)}{b - a} \][/tex]
For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = -\left(1 - \frac{7}{2}\right)^2 + \frac{81}{4} \][/tex]
[tex]\[ f(1) = -\left(-\frac{5}{2}\right)^2 + \frac{81}{4} \][/tex]
[tex]\[ f(1) = -\left(\frac{25}{4}\right) + \frac{81}{4} \][/tex]
[tex]\[ f(1) = -\frac{25}{4} + \frac{81}{4} = \frac{56}{4} = 14 \][/tex]
For [tex]\( x = 5 \)[/tex]:
[tex]\[ f(5) = -\left(5 - \frac{7}{2}\right)^2 + \frac{81}{4} \][/tex]
[tex]\[ f(5) = -\left(\frac{3}{2}\right)^2 + \frac{81}{4} \][/tex]
[tex]\[ f(5) = -\frac{9}{4} + \frac{81}{4} = \frac{72}{4} = 18 \][/tex]
Now, average gradient:
[tex]\[ \text{Average Gradient} = \frac{f(5) - f(1)}{5-1} = \frac{18 - 14}{4} = \frac{4}{4} = 1 \][/tex]
### 5.3 Calculate the [tex]\( x \)[/tex]-coordinate of Point [tex]\( D \)[/tex]
To find the intersection points of [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex]:
[tex]\[ f(x) = g(x) \][/tex]
[tex]\[ -\left(x-\frac{7}{2}\right)^2 + \frac{81}{4} = -3x + 24 \][/tex]
First, simplify:
[tex]\[ \left(x-\frac{7}{2}\right)^2 = 3x - 24 + \frac{81}{4} \][/tex]
[tex]\[ \left(x-\frac{7}{2}\right)^2 = 3x - 24 + 20.25 \][/tex]
[tex]\[ \left(x-\frac{7}{2}\right)^2 = 3x - 3.75 \][/tex]
Simplify into the quadratic equation:
[tex]\[ \left(x - \frac{7}{2}\right)^2 - 3x + 3.75 = 0 \][/tex]
[tex]\[ x^2 - 7x + \frac{49}{4} - 3x + 3.75 = 0 \][/tex]
[tex]\[ x^2 - 10x + \frac{49}{4} + 3.75 = 0 \][/tex]
[tex]\[ x^2 - 10x + \frac{64}{4} = 0 \][/tex]
[tex]\[ x^2 - 10x + 16 = 0 \][/tex]
Solve using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ a = 1, b = -10, c = 16 \][/tex]
[tex]\[ x = \frac{10 \pm \sqrt{100 - 64}}{2} \][/tex]
[tex]\[ x = \frac{10 \pm \sqrt{36}}{2} \][/tex]
[tex]\[ x = \frac{10 \pm 6}{2} \][/tex]
[tex]\[ x = 8 \quad \text{or} \quad x = 2 \][/tex]
Assuming [tex]\( D \)[/tex] has the higher [tex]\( x \)[/tex] value, [tex]\( a = 8 \)[/tex].
### 5.4 Determine [tex]\( ST \)[/tex] in terms of [tex]\( x \)[/tex]
[tex]\( S(x, y) \)[/tex] is a point on the graph of [tex]\( f \)[/tex], and [tex]\( T(x, y) \)[/tex] is a point on the graph of [tex]\( g \)[/tex]:
[tex]\[ ST = f(x) - g(x) \][/tex]
Given:
[tex]\[ f(x) = -\left(x - \frac{7}{2}\right)^2 + \frac{81}{4} \][/tex]
[tex]\[ g(x) = -3x + 24 \][/tex]
Thus:
[tex]\[ ST = -\left(x - \frac{7}{2}\right)^2 + \frac{81}{4} + 3x - 24 \][/tex]
### 5.5 Calculate the Maximum Length of [tex]\( ST \)[/tex]
To find the maximum length of [tex]\( ST \)[/tex]:
[tex]\[ ST = -\left(x - \frac{7}{2}\right)^2 + \frac{81}{4} + 3x - 24 \][/tex]
Find the critical points by differentiating and setting the derivative to zero:
[tex]\[ \frac{d}{dx} \left[-\left(x - \frac{7}{2}\right)^2 + \frac{81}{4} + 3x - 24\right] = 0 \][/tex]
[tex]\[ \frac{d}{dx} \left[ -\left(x^2 - 7x + \frac{49}{4}\right) + \frac{81}{4} + 3x - 24 \right] = 0 \][/tex]
[tex]\[ -2 (x - \frac{7}{2}) + 3 = 0 \][/tex]
[tex]\[ -2x + 7 + 3 = 0 \][/tex]
[tex]\[ -2x + 10 = 0 \][/tex]
[tex]\[ x = 5 \][/tex]
Therefore:
[tex]\[ ST_{\max} = f(5) - g(5) \][/tex]
Evaluate:
[tex]\[ f(5) = 18 \][/tex]
[tex]\[ g(5) = -15 + 24 = 9 \][/tex]
[tex]\[ ST_{\max} = 18 - 9 = 9 \][/tex]
So the maximum length of [tex]\( ST \)[/tex] is [tex]\( 9 \)[/tex].
### 5.1 Coordinates of [tex]\( E \)[/tex]
The turning point of a quadratic function in the form [tex]\( f(x) = a(x - h)^2 + k \)[/tex] is at the point [tex]\((h, k)\)[/tex]. Given:
[tex]\[ f(x) = -\left(x - \frac{7}{2}\right)^2 + \frac{81}{4} \][/tex]
The vertex, which is the turning point [tex]\( E \)[/tex], has the coordinates:
[tex]\[ E \left( \frac{7}{2}, \frac{81}{4} \right) \][/tex]
### 5.2 Average Gradient of [tex]\( f \)[/tex] between [tex]\( x = 1 \)[/tex] and [tex]\( x = 5 \)[/tex]
The average gradient of a function between two points [tex]\( x=a \)[/tex] and [tex]\( x=b \)[/tex] is calculated as:
[tex]\[ \text{Average Gradient} = \frac{f(b) - f(a)}{b - a} \][/tex]
For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = -\left(1 - \frac{7}{2}\right)^2 + \frac{81}{4} \][/tex]
[tex]\[ f(1) = -\left(-\frac{5}{2}\right)^2 + \frac{81}{4} \][/tex]
[tex]\[ f(1) = -\left(\frac{25}{4}\right) + \frac{81}{4} \][/tex]
[tex]\[ f(1) = -\frac{25}{4} + \frac{81}{4} = \frac{56}{4} = 14 \][/tex]
For [tex]\( x = 5 \)[/tex]:
[tex]\[ f(5) = -\left(5 - \frac{7}{2}\right)^2 + \frac{81}{4} \][/tex]
[tex]\[ f(5) = -\left(\frac{3}{2}\right)^2 + \frac{81}{4} \][/tex]
[tex]\[ f(5) = -\frac{9}{4} + \frac{81}{4} = \frac{72}{4} = 18 \][/tex]
Now, average gradient:
[tex]\[ \text{Average Gradient} = \frac{f(5) - f(1)}{5-1} = \frac{18 - 14}{4} = \frac{4}{4} = 1 \][/tex]
### 5.3 Calculate the [tex]\( x \)[/tex]-coordinate of Point [tex]\( D \)[/tex]
To find the intersection points of [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex]:
[tex]\[ f(x) = g(x) \][/tex]
[tex]\[ -\left(x-\frac{7}{2}\right)^2 + \frac{81}{4} = -3x + 24 \][/tex]
First, simplify:
[tex]\[ \left(x-\frac{7}{2}\right)^2 = 3x - 24 + \frac{81}{4} \][/tex]
[tex]\[ \left(x-\frac{7}{2}\right)^2 = 3x - 24 + 20.25 \][/tex]
[tex]\[ \left(x-\frac{7}{2}\right)^2 = 3x - 3.75 \][/tex]
Simplify into the quadratic equation:
[tex]\[ \left(x - \frac{7}{2}\right)^2 - 3x + 3.75 = 0 \][/tex]
[tex]\[ x^2 - 7x + \frac{49}{4} - 3x + 3.75 = 0 \][/tex]
[tex]\[ x^2 - 10x + \frac{49}{4} + 3.75 = 0 \][/tex]
[tex]\[ x^2 - 10x + \frac{64}{4} = 0 \][/tex]
[tex]\[ x^2 - 10x + 16 = 0 \][/tex]
Solve using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ a = 1, b = -10, c = 16 \][/tex]
[tex]\[ x = \frac{10 \pm \sqrt{100 - 64}}{2} \][/tex]
[tex]\[ x = \frac{10 \pm \sqrt{36}}{2} \][/tex]
[tex]\[ x = \frac{10 \pm 6}{2} \][/tex]
[tex]\[ x = 8 \quad \text{or} \quad x = 2 \][/tex]
Assuming [tex]\( D \)[/tex] has the higher [tex]\( x \)[/tex] value, [tex]\( a = 8 \)[/tex].
### 5.4 Determine [tex]\( ST \)[/tex] in terms of [tex]\( x \)[/tex]
[tex]\( S(x, y) \)[/tex] is a point on the graph of [tex]\( f \)[/tex], and [tex]\( T(x, y) \)[/tex] is a point on the graph of [tex]\( g \)[/tex]:
[tex]\[ ST = f(x) - g(x) \][/tex]
Given:
[tex]\[ f(x) = -\left(x - \frac{7}{2}\right)^2 + \frac{81}{4} \][/tex]
[tex]\[ g(x) = -3x + 24 \][/tex]
Thus:
[tex]\[ ST = -\left(x - \frac{7}{2}\right)^2 + \frac{81}{4} + 3x - 24 \][/tex]
### 5.5 Calculate the Maximum Length of [tex]\( ST \)[/tex]
To find the maximum length of [tex]\( ST \)[/tex]:
[tex]\[ ST = -\left(x - \frac{7}{2}\right)^2 + \frac{81}{4} + 3x - 24 \][/tex]
Find the critical points by differentiating and setting the derivative to zero:
[tex]\[ \frac{d}{dx} \left[-\left(x - \frac{7}{2}\right)^2 + \frac{81}{4} + 3x - 24\right] = 0 \][/tex]
[tex]\[ \frac{d}{dx} \left[ -\left(x^2 - 7x + \frac{49}{4}\right) + \frac{81}{4} + 3x - 24 \right] = 0 \][/tex]
[tex]\[ -2 (x - \frac{7}{2}) + 3 = 0 \][/tex]
[tex]\[ -2x + 7 + 3 = 0 \][/tex]
[tex]\[ -2x + 10 = 0 \][/tex]
[tex]\[ x = 5 \][/tex]
Therefore:
[tex]\[ ST_{\max} = f(5) - g(5) \][/tex]
Evaluate:
[tex]\[ f(5) = 18 \][/tex]
[tex]\[ g(5) = -15 + 24 = 9 \][/tex]
[tex]\[ ST_{\max} = 18 - 9 = 9 \][/tex]
So the maximum length of [tex]\( ST \)[/tex] is [tex]\( 9 \)[/tex].
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