Question

I require some assistance with this graphing question, please.

"Use the parabola tool to graph the quadratic function
f(x)=−(x+3)^2+5

Graph the parabola by first plotting its vertex and then plotting a second point on the parabola."

The graph's max on both the X and Y axis is 10, and goes no further.
Any help would be appreciated, but feel free to take your time.

Answer 1

y-int = 5

roots: sqrt(5)-3 or -3 - sqrt(5)

TP @ (-3,5)

Explanation:

y intercept = 5 (when x = 0)

Roots:

When y = 0

5 - (x + 3)^2 = 0

(x+3)^2 = 5

Square both sides:

x + 3 = Sqrt[5] or x + 3 = - Sqrt[5]

x = Sqrt[5] - 3 or x= - 3 - Sqrt[5]

Turning point (Critical Point):

dy/dx (5-(x+3)^2) = - 2 (x+3)

Solve -2 (x+3) = 0

x = - 3

y = 5

Max point at (-3,5)

Answer 2

vertex (-3,5) and another pt (-2,4)

Explanation:

It is in vertex form so the vertex is (-3,5)...

Now just plug in a value for x say like -2...

f(-2)=-(-2+3)^2+5

f(-2)=-(1)^2+5

f(-2)=-1+5

f(-2)=4

So another point is (-2,4)