Question

For each of the following problems: a) State the Parent function b) Graph the equation c) Indently the transformationsd) State the DOMAIN and RANGE 14) y = -(x+3) ² - 315) y = 1/2 -(x -4)^2 +3

Answer 1

14) We have the function:

$-(x+3)^2-3$

a) The parent function is the quadratic parent function f(x) = x^2.

b) We can graph the equation knowing that the vertex is at x=-3. The corresponding value for x=-3 is:

$f(-3)=-(-3+3)^2-3=-0-3=-3$

The vertex is (-3,-3).

As the sign in the quadratic term is neative, the parabola is concave down.

c) The transformation can be listed as:

- Reflection over the x-axis: this converts the original concave-up parabola into a concave-down parabola:

$\begin{gathered} (x,y)\longrightarrow(x,-y) \\ x^2\longrightarrow-x^2 \end{gathered}$

- Displacement 3 units to the left and 3 units down.

$\begin{gathered} (x,-y)\longrightarrow(x-(-3),-y-3)=(x+3,-y-3) \\ -x^2\longrightarrow-(x+3)^2-3 \end{gathered}$

d) The domain is all the real numbers as there is no value of x for which f(x) is not defined.

The range can be defined as all the values of y that are equal or less than -3 (y<=-3).

15) We have the function:

$f(x)=\frac{1}{2}-(x-4)^2+3=-(x-4)^2+\frac{6}{2}+\frac{1}{2}=-(x-4)^2+\frac{7}{2}$

This case is similar to the previous one, but with a different displacement.

a) The parent function is the quadratic parent function f(x)=x^2.

b) The graph is:

The vertex is now (4,7/2).

The negative sign in the quadratic term tells us that the parabola is concave down.

c) The transformations are:

- Reflection over the x-axis.

- Displacement 4 units right and 7/2 units up.

d) The domain is all the real values and the range is all teh values of y that are less or equal than 7/2 (y<=7/2).