Answer:
Point A: (2, 10)
Point B: (-3, 0)
Point C: (-5, -4)
Point D: (-5, -32)
Step-by-step explanation:
Points A and B are the points of intersection between the two graphs.
Therefore, to find the x-values of the points of intersection, substitute one equation into the other and solve for x:
[tex]\implies 2x+6=-2x^2+18[/tex]
[tex]\implies 2x^2+2x-12=0[/tex]
[tex]\implies 2(x^2+x-6)=0[/tex]
[tex]\implies x^2+x-6=0[/tex]
[tex]\implies x^2+3x-2x-6=0[/tex]
[tex]\implies x(x+3)-2(x+3)=0[/tex]
[tex]\implies (x-2)(x+3)=0[/tex]
[tex]\implies x=2, -3[/tex]
From inspection of the graph:
To find the y-values, substitute the found x-values into either of the equations:
[tex]\begin{aligned} \textsf{Point A}: \quad 2x+6 & =y\\2(2)+6 & =10\\ \implies & (2, 10)\end{aligned}[/tex]
[tex]\begin{aligned} \textsf{Point B}: \quad -2x^2+18 & =y\\-2(-3)^2+18 & =0\\ \implies & (-3,0)\end{aligned}[/tex]
Therefore, point A is (2, 10) and point B is (-3, 0).
If the distance between points C and D is 28 units, the y-value of point D will be 28 less than the y-value of point C. The x-values of the two points are the same.
Therefore:
[tex]\textsf{Equation 1}: \quad y=2x+6[/tex]
[tex]\textsf{Equation 2}: \quad y-28=-2x^2+18[/tex]
As the x-values are the same, substitute the first equation into the second equation and solve for x to find the x-value of points C and D:
[tex]\implies 2x+6-28=-2x^2+18[/tex]
[tex]\implies 2x^2+2x-40=0[/tex]
[tex]\implies 2(x^2+x-20)=0[/tex]
[tex]\implies x^2+x-20=0[/tex]
[tex]\implies x^2+5x-4x-20=0[/tex]
[tex]\implies x(x+5)-4(x+5)=0[/tex]
[tex]\implies (x-4)(x+5)=0[/tex]
[tex]\implies x=4,-5[/tex]
From inspection of the given graph, the x-value of points C and D is negative, therefore x = -5.
To find the y-value of points C and D, substitute the found value of x into the two original equations of the lines:
[tex]\begin{aligned} \textsf{Point C}: \quad 2x+6 & =y\\2(-5)+6 & =-4\\ \implies & (-5,-4)\end{aligned}[/tex]
[tex]\begin{aligned} \textsf{Point D}: \quad -2x^2+18 & = y \\ -2(-5)^2+18 & =-32\\ \implies & (-5, -32)\end{aligned}[/tex]
Therefore, point C is (-5, -4) and point D is (-5, -32).
Answer:
a) A = (2, 10) and B = (-3, 0)
b) C = (-5, -4) and D = (-5, -32)
Explanation:
a) To determine the coordinates of A and B, find the intersection points of the line "y = 2x + 6" and curve "y = -2x² + 18".
Solve the equation's simultaneously:
y = y
⇒ 2x + 6 = -2x² + 18
⇒ 2x² + 2x + 6 - 18 = 0
⇒ 2x² + 2x - 12= 0
⇒ 2x² + 6x - 4x - 12 = 0
⇒ 2x(x + 3) - 4(x + 3) = 0
⇒ (2x - 4)(x + 3) = 0
⇒ 2x - 4 = 0, x + 3 = 0
⇒ x = 2, x = -3
Then find value of y at this x points,
at x = 2, y = 2(2) + 6 = 10
at x = -3, y = 2(-3) + 6 = 0
Intersection points: A(2, 10) and B(-3, 0)
b) Given that CD = 28 units. Also stated parallel to y axis so x coordinates for both will be same but differ in y coordinate.
[tex]2x + 6 = -2x^2 + 18 + 28[/tex]
[tex]-2x^2 + 18 + 28-2x - 6 = 0[/tex]
[tex]-2x^2-2x+40=0[/tex]
[tex]-2x^2-10x+8x+40=0[/tex]
[tex]-2(x+5)+8(x+5)=0[/tex]
[tex](-2x+8)(x+5)=0[/tex]
[tex]x = -5, 4[/tex]
[tex]\leftrightarrow \sf C(-5, y_2), \ D(-5, y_2)[/tex]
Find y value for Point C : 2x + 6 = 2(-5) + 6 = -4
Find y value for Point D : -2x² + 18 = -2(-5)² + 18 = -32
[tex]\sf \rightarrow Point \ C = (-5, -4)\\ \\\rightarrow Point \ D = (-5, -32)[/tex]