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:
x= 3
Step-by-step explanation:
1. -5(3) + 2 = -13
2. -15 + 2 = -13 <em>A positive times a negative will always equal a negative</em>
3. -13 = -13 <em>Whenever you add anything to a negative number, the </em>
<em> number becomes smaller </em>
Ex: -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, ...........
<em>Although the number itself becomes smaller, its value </em>
<em> becomes bigger </em>
Hope this helps.
Answer:
Step-by-step explanation:
Quadratic equations of the form (x - h)^2 + k = 0 can be solved using square roots: Take the square root of both sides, prefacing the right side with " ± "
