Find the coordinates of the vertex for the parabola defined by the given quadratic function f(x) = (x + 2)^2 + 2 a) (0,2) b) (2.0) c) (-2.2) d) (-2,-2) e) none

(-2,2)

Step-by-step explanation:

Let's find the answer.

Because a tangent line for a parabola function is equal to 0 only at its vertex then:

so then:

when

For x=-2 f(x) is:

In conclusion, the vertex of the given parabola is (-2,2), so the answer is C. Although in your answer is reported as (-2.2) but I think was a typing mistake.

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3