Completing the square has us breaking rules of solving equations and factoring out the greatest common factor, but it is what it is! The first step is to make sure that the coefficient on the x^2 term is a 1 and it is so we are good there. Now subtract 9 from both sides to get x^2 + 16x = -9. Complete the square on the left side by taking half of the linear term (16x) which is 8 and then squaring it to get 64. That's what is added to both sides. Now it looks like this:
x^2 + 16x + 64 = -9 + 64. If you were to write it in vertex form it would look like this: (x+8)^2 - 55 = 0. Now you can use this to plot the vertex of a parabola if you want to: it sits at (-8, -55)
Answer:
Anika is correct
Step-by-step explanation:
* Lets revise the rules of reflection and rotation
- If point (x , y) reflected across the x-axis then its image is (x , -y)
- If point (x , y) rotated about the origin by angle 270° clock wise then
its image is (-y , x)
* Lets check the vertices of the two triangles
- The vertices of Δ ABC are:
# A (-7 , 2)
# B (-3 , 6)
# C (-2 , 1)
- The vertices of Δ PRQ
# P (-7 , -2)
# R (-3 , -6)
# Q (-2 , -1)
* By comparing between the vertices of the two triangles
∵ Each y-coordinates of Δ PRQ has opposite sign of each
y-coordinates of Δ ABC
∵ All x-coordinates of Δ PRQ are the same with x-coordinates of
Δ ABC
- That means Δ PRQ is the image of Δ ABC after reflection across
the x-axis
∵ Reflection doesn't change the shape and the size of the figure
∴ Δ ABC and Δ PRQ have same size (equal sides and equal angles)
∴ Δ ABC is congruent to Δ PRQ
* Anika is correct