Answer:
3) (2,-9)
4) (0,-5)
5) (1,-8)
Step-by-step explanation:
3)
The vertex will occur between you x-intercepts.
You already found that happens at x=2.
To find the corresponding y-coordinate, replace x in
f(x)=(x+1)(x-5) with 2:
f(2)=(2+1)(2-5)
f(2)=(3)(-3)
f(2)=-9
So the vertex is (2,-9).
4)
The y-intercept is when x=0.
So in f(x)=(x+1)(x-5) replace x with 0:
f(0)=(0+1)(0-5)
f(0)=(1)(-5)
f(0)=-5
So the y-intercept is (0,-5).
5)
To find another point just plug in anything besides any x already used.
We preferably want to use a value of x that will keep us on their grid however far up,down,left, or right their grid goes out. So I'm going to choose something close to the vertex which is at x=2. Let's go with x=1.
So replace x in f(x)=(x+1)(x-5) with x=1:
f(1)=(1+1)(1-5)
f(1)=(2)(-4)
f(1)=-8
So another point to graph is (1,-8).
Answer:
principal is the correct answer
Step-by-step explanation:
principal is the correct answer
Answer:
second option
Step-by-step explanation:
The equation of a circle in standard form is
(x - h)² + (y - k)² = r²
where (h, k) are the coordinates of the centre and r is the radius
Here (h, k) = (4, 5) and r = 2 , thus
(x - 4)² + (y - 5)² = 2², that is
(x - 4)² + (y - 5)² = 4 ← equation of circle
Answer:
Step-by-step explanation:
Put n = 2 and next n = 3 to the recursive formula:
D the rest would be impossible to recreate
