Answer:
Point C: (3, 3)
Point D: (3, -22)
Step-by-step explanation:
If the distance between points C and D is 25 units, the y-value of point D will be 25 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-3[/tex]
[tex]\textsf{Equation 2}: \quad y-25=-3x^2+5[/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-3-25=-3x^2+5[/tex]
[tex]\implies 3x^2+2x-33=0[/tex]
[tex]\implies 3x^2-9x+11x-33=0[/tex]
[tex]\implies 3x(x-3)+11(x-3)=0[/tex]
[tex]\implies (x-3)(3x+11)=0[/tex]
[tex]\implies x=3, -\dfrac{11}{3}[/tex]
From inspection of the given graph, the x-value of points C and D is positive, therefore x = 3.
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-3 & =y\\2(3)-3 & =3\\ \implies & (3, 3)\end{aligned}[/tex]
[tex]\begin{aligned} \textsf{Point D}: \quad -3x^2+5 & = y \\ -3(3)^2+5 & =-22\\ \implies & (3, -22)\end{aligned}[/tex]
Therefore, point C is (3, 3) and point D is (3, -22).
