Put Y=x^2+10x+18 in vertex form
2 answers:
You might be interested in
Answer:
Please find attached the required plot accomplished with an online tool
Part A:
1/4
Part B:
P''(-1, 0), Q''(0, -1), and R''(2, -1)
Part C:
Triangle PQR is similar to triangle P''Q''R'' but they are not congruent
Step-by-step explanation:
Part A:
Triangle ΔPQR has vertices P(4, 0), Q(0, -4), R(-8, -4)
Triangle ΔP'Q'R' has vertices P'(1, 0), Q'(0, -1), R'(-2, -1)
The dimensions of the sides of the triangle are given by the relation;
Where;
(x₁, y₁) and (x₂, y₂) are the coordinates on the ends of the segment
For segment PQ, we place (x₁, y₁) = (4, 0) and (x₂, y₂) = (0, -4);
By substitution into the length equation, we get;
The length of segment PQ = 4·√2
The length of segment PR = 4·√10
The length of segment RQ = 8
The length of segment P'Q' = √2
The length of segment P'R' = √10
The length of segment R'Q' = 2
Therefore, the scale factor of the dilation of ΔPQR to ΔP'Q'R' is 1/4
Part B:
Reflection of (x, y) across the y-axis gives;
(x, y) image after reflection across the y-axis = (-x, y)
The coordinates after reflection of P'(1, 0), Q'(0, -1), R'(-2, -1) across the y-axis is given as follows;
P'(1, 0) image after reflection across the y-axis = P''(-1, 0)
Q'(0, -1) image after reflection across the y-axis = Q''(0, -1)
R'(-2, -1) image after reflection across the y-axis = R''(2, -1)
Part C:
Triangle PQR is similar to triangle P''Q''R'' but they are not congruent as the dimensions of the sides of triangle PQR and P''Q''R'' are not the same.
