Plug x = 0 into the function
f(x) = x^3 + 2x - 1
f(0) = 0^3 + 2(0) - 1
f(0) = -1
Note how the result is negative. The actual number itself doesn't matter. All we care about is the sign of the result.
Repeat for x = 1
f(x) = x^3 + 2x - 1
f(1) = 1^3 + 2(1) - 1
f(1) = 2
This result is positive.
---------------------------
We found that f(0) = -1 and f(1) = 2. The first output -1 is negative while the second output 2 is positive. Going from negative to positive means that, at some point, we will hit y = 0. We might have multiple instances of this happening, or just one. We don't know for sure. The only thing we do know is that there is at least one root in this interval.
To actually find this root, you'll need to use a graphing calculator because the root is some complicated decimal value. Using a graphing calculator, you should find the root to be approximately 0.4533976515
Sorry but grade math is this so I can have a better understanding
A. d=50t B. 50,100,150,200,250 C. 50 days
Given:
The graph of line.
To find:
The gradient of the line using rise/run method.
Solution:
We know that the gradient of a line is also known as slope.
Consider the two intercepts, then rise is distance between origin and y-intercept and run is the distance between origin and the x-intercept. But rise must be negative because the value of y decreased from 3 to 0.
Now,
The gradient of the line is 
.
Therefore, the correct option is C.
Answer: y - 5 = 0(x - 1)
==================================================
Explanation:
Recall that point slope form in general is written as such
y - y1 = m(x - x1)
where,
m is the slope
(x1,y1) is the point the line goes through
The given equation y = 7 can be written as y = 0x+7. So we see that this line has a slope of m = 0
Plug m = 0 along with the given point (x1,y1) = (1,5) into the point slope equation and we get
y - y1 = m(x - x1)
y - 5 = 0(x - 1)
which is the final answer
note: the equation in bold can be rearranged and simplified to get y = 5; however your teacher seems to want the answer in point-slope form, so we leave it as such.
