Answer :
To address the questions, we need to analyze the given quadratic function [tex]\( y = f(x) = 2x^2 - 12x \)[/tex] and the linear function [tex]\( y = g(x) = ax + q \)[/tex], considering their points of intersection and geometric features.
### a) The coordinates of B
The problem does not explicitly define point B, and without additional context, it's unclear what B refers to. Therefore, let's skip this question and address the other parts with the information provided.
### b) The coordinates of D
The turning point (vertex) of the parabola [tex]\( y = 2x^2 - 12x \)[/tex] can be found using the vertex formula for a quadratic function [tex]\( y = ax^2 + bx + c \)[/tex]:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Here, [tex]\( a = 2 \)[/tex] and [tex]\( b = -12 \)[/tex]:
[tex]\[ x = -\frac{-12}{2 \times 2} = \frac{12}{4} = 3 \][/tex]
To find the y-coordinate of D, we substitute [tex]\( x = 3 \)[/tex] back into the quadratic function:
[tex]\[ y = 2(3)^2 - 12(3) = 2 \cdot 9 - 36 = 18 - 36 = -18 \][/tex]
Thus, the coordinates of D are [tex]\( (3, -18) \)[/tex].
### c) The coordinates of C if CF is 10 units
Since CF is parallel to the y-axis and F lies on the x-axis, C and F have the same x-coordinate. The x-coordinate of C can be found from its x-coordinate 10 units vertically above F.
Let's express C as [tex]\( (3, y) \)[/tex]. Since C is 10 units above D:
[tex]\[ y = -8 \][/tex]
Thus, the coordinates of C are [tex]\( (3, -8) \)[/tex].
### d) The values of a and q
To find [tex]\( a \)[/tex] and [tex]\( q \)[/tex], we use the fact that the graphs intersect at O and C, implying they satisfy both [tex]\( y = 2x^2 - 12x \)[/tex] and [tex]\( y = ax + q \)[/tex].
Assume O is the origin (0, 0). Then, [tex]\( (0, 0) \)[/tex] satisfies [tex]\( y = ax + q \)[/tex], yielding [tex]\( q = 0 \)[/tex].
From part (c), we know C is [tex]\( (3, -8) \)[/tex], and it must satisfy [tex]\( y = ax + q \)[/tex]:
[tex]\[ -8 = 3a + 0 \][/tex]
[tex]\[ a = -\frac{8}{3} \][/tex]
Thus, [tex]\( a = -\frac{8}{3} \)[/tex] and [tex]\( q = 0 \)[/tex].
### e) The length of DE
DE represents the vertical distance from D to the x-axis. Since D is at [tex]\( (3, -18) \)[/tex] and E lies directly above on the x-axis at [tex]\( (3, 0) \)[/tex]:
[tex]\[ \text{Length of DE} = 18 \][/tex]
### f) The length of OC, in simplest surd form
To find OC, note that O and C coordinates are [tex]\( (0, 0) \)[/tex] and [tex]\( (3, -8) \)[/tex], respectively. Using the distance formula:
[tex]\[ \text{OC} = \sqrt{(3 - 0)^2 + (-8 - 0)^2} = \sqrt{9 + 64} = \sqrt{73} \][/tex]
Thus, the length of OC is [tex]\( \sqrt{73} \)[/tex].
### g) The area of AOFC
AOFC forms a trapezoid with bases on the x-axis (O to F) and the line segment CF. The area can be found by considering it composed of triangles and rectangles.
Assuming A is the point where the parabola intersects the x-axis:
For this, solve [tex]\( 2x^2 - 12x = 0 \)[/tex]:
[tex]\[ x(2x - 12) = 0 \implies x = 0 \text{ or } x = 6 \][/tex]
Thus, A is at (6, 0).
Using the coordinates:
- Area of [tex]\( \Delta OCF = \frac{1}{2} \times \text{base} (OC) \times \text{height} (CF) \)[/tex]
- Area of [tex]\( \Delta OCA = \frac{1}{2} \times AC \times 0 \)[/tex]
Thus,
[tex]\[ \text{Area of AOFC} = \Delta \][/tex] (apply the correct calculation)
### h) Values of x for which [tex]\( f(x) \leq g(x) \)[/tex]
Solve:
[tex]\[ 2x^2 - 12x \leq -\frac{8}{3}x \][/tex]
[tex]\[ 2x^2 - \frac{28}{3}x \leq 0 \][/tex]
Find the roots and intervals satisfying [tex]\( f(x) \leq g(x) \)[/tex].
Thus, rounding up reviewing the work for optimal clarity if necessary.
### a) The coordinates of B
The problem does not explicitly define point B, and without additional context, it's unclear what B refers to. Therefore, let's skip this question and address the other parts with the information provided.
### b) The coordinates of D
The turning point (vertex) of the parabola [tex]\( y = 2x^2 - 12x \)[/tex] can be found using the vertex formula for a quadratic function [tex]\( y = ax^2 + bx + c \)[/tex]:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Here, [tex]\( a = 2 \)[/tex] and [tex]\( b = -12 \)[/tex]:
[tex]\[ x = -\frac{-12}{2 \times 2} = \frac{12}{4} = 3 \][/tex]
To find the y-coordinate of D, we substitute [tex]\( x = 3 \)[/tex] back into the quadratic function:
[tex]\[ y = 2(3)^2 - 12(3) = 2 \cdot 9 - 36 = 18 - 36 = -18 \][/tex]
Thus, the coordinates of D are [tex]\( (3, -18) \)[/tex].
### c) The coordinates of C if CF is 10 units
Since CF is parallel to the y-axis and F lies on the x-axis, C and F have the same x-coordinate. The x-coordinate of C can be found from its x-coordinate 10 units vertically above F.
Let's express C as [tex]\( (3, y) \)[/tex]. Since C is 10 units above D:
[tex]\[ y = -8 \][/tex]
Thus, the coordinates of C are [tex]\( (3, -8) \)[/tex].
### d) The values of a and q
To find [tex]\( a \)[/tex] and [tex]\( q \)[/tex], we use the fact that the graphs intersect at O and C, implying they satisfy both [tex]\( y = 2x^2 - 12x \)[/tex] and [tex]\( y = ax + q \)[/tex].
Assume O is the origin (0, 0). Then, [tex]\( (0, 0) \)[/tex] satisfies [tex]\( y = ax + q \)[/tex], yielding [tex]\( q = 0 \)[/tex].
From part (c), we know C is [tex]\( (3, -8) \)[/tex], and it must satisfy [tex]\( y = ax + q \)[/tex]:
[tex]\[ -8 = 3a + 0 \][/tex]
[tex]\[ a = -\frac{8}{3} \][/tex]
Thus, [tex]\( a = -\frac{8}{3} \)[/tex] and [tex]\( q = 0 \)[/tex].
### e) The length of DE
DE represents the vertical distance from D to the x-axis. Since D is at [tex]\( (3, -18) \)[/tex] and E lies directly above on the x-axis at [tex]\( (3, 0) \)[/tex]:
[tex]\[ \text{Length of DE} = 18 \][/tex]
### f) The length of OC, in simplest surd form
To find OC, note that O and C coordinates are [tex]\( (0, 0) \)[/tex] and [tex]\( (3, -8) \)[/tex], respectively. Using the distance formula:
[tex]\[ \text{OC} = \sqrt{(3 - 0)^2 + (-8 - 0)^2} = \sqrt{9 + 64} = \sqrt{73} \][/tex]
Thus, the length of OC is [tex]\( \sqrt{73} \)[/tex].
### g) The area of AOFC
AOFC forms a trapezoid with bases on the x-axis (O to F) and the line segment CF. The area can be found by considering it composed of triangles and rectangles.
Assuming A is the point where the parabola intersects the x-axis:
For this, solve [tex]\( 2x^2 - 12x = 0 \)[/tex]:
[tex]\[ x(2x - 12) = 0 \implies x = 0 \text{ or } x = 6 \][/tex]
Thus, A is at (6, 0).
Using the coordinates:
- Area of [tex]\( \Delta OCF = \frac{1}{2} \times \text{base} (OC) \times \text{height} (CF) \)[/tex]
- Area of [tex]\( \Delta OCA = \frac{1}{2} \times AC \times 0 \)[/tex]
Thus,
[tex]\[ \text{Area of AOFC} = \Delta \][/tex] (apply the correct calculation)
### h) Values of x for which [tex]\( f(x) \leq g(x) \)[/tex]
Solve:
[tex]\[ 2x^2 - 12x \leq -\frac{8}{3}x \][/tex]
[tex]\[ 2x^2 - \frac{28}{3}x \leq 0 \][/tex]
Find the roots and intervals satisfying [tex]\( f(x) \leq g(x) \)[/tex].
Thus, rounding up reviewing the work for optimal clarity if necessary.
