Answer:
Step-by-step explanation:
f"(x)=2
integrating
f'(x)=2x+c
f'(1)=2+c=4
c=4-2=2
f'(x)=2x+2
integrating
f(x)=2x^2/2+2x+a
f(x)=x^2+2x+a
f(2)=-2
(2)^2+2(2)+a=-2
4+4+a=-2
a=-2-8=-10
f(x)=x^2+2x-10
Hello! In slope-intercept form, y=mx+b. M is the slope and b is the y-intercept. The rate of change and slope are interchangeable, therefore, they both create a line through the slope-intercept form.
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Option A. All the real values of x where x < -1
Procedure
Solve the inequality:
(x -3)(x+1)>0
That happens in two cases.
1) When both factors >0
x-3>0 and x+1>0
x>3 and x >-1
The intersection is x >3
2) When both factors <0
x-3<0 and x+1<0
x<3 and x<-1
the intersection is x<-1.
We have obtained that the function is positive for the intervals x < -1 and x > 3. But in one of those intervals the function is decresing and in the other is increasing.
You can recognize that the function given is a parabola and, because the coefficient of the quadratic term is positive, the parabola opens upward. Then the function is decreasing in the first interval and increasing in the second interval.
<h2>
Answer:</h2>
If the line is parallel to the one given, their slopes are the same, so we need to find the y-intercept, then we can determine the slope-intercept form parallel to the given line.
Now that we have the y-intercept, we can create the equation.
We can prove this by graphing both lines, like so.
