87.0k views
0 votes
Use y = x^2 + 12x + 11(-2, -9) is a point on the graph. What point is the reflection of (-2,-9) across the axis of symmetry of the parabola?

1 Answer

0 votes

The equation:


y=x^2+12x+11

has the form:


y=ax^2+bx+c

with a = 1, b = 12 and c = 11.

The x-coordinate of the vertex, Xv, is found as follows:


\begin{gathered} x_v=(-b)/(2a)_{} \\ x_v=(-12)/(2\cdot1) \\ x_v=(-12)/(2) \\ x_v=-6 \end{gathered}

And the equation of the axis of symmetry of the parabola is:


\begin{gathered} x=x_v \\ x=-6 \end{gathered}

The point (-2, -9) is located 4 units to the right of the axis of symmetry. Then, the reflected point must be located 4 units to the left, its x-coordinate must be:

x = -6 - 4 = -10

Evaluating this point into the function:


\begin{gathered} y=(-10)^2+12\cdot(-10)+11 \\ y=100+-120+11 \\ y=-9 \end{gathered}

Therefore, the reflected point is (-10, -9).

User Seshu Vuggina
by
7.7k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories