I require some assistance with this graphing question, please.
2 answers:
Answer:
y-int = 5
roots: sqrt(5)-3 or -3 - sqrt(5)
TP @ (-3,5)
Step-by-step 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)
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Answer:
Step-by-step explanation:
Vertex form is accomplished by completing the square on the quadratic. Do this by first setting the parabola equal to 0 then moving the constant over to the other side:
Now take half the linear term, square it, and add it to both sides. Our linear term is 6. Half of 6 is 3, and 3 squared is 9:
The reason we do this is to create a perfect square binomial on the left:
(obviously the 0 results from the addition of 9 and -9). Move the 0 back over to the other side and set the quadratic back equal to y:
This gives you a vertex of (-3, 0), which is a minimum value, since the parabola opens upwards.