Find the vertex of y=-5x^2+10x+3
2 answers:
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Answer:
The set fee would be $15
Explanation:
The set fee is the starting value. This means that it is the value of the y at x = 0 (y-intercept).
To get the set fee, we would first need to get the equation of the line.
Equation of the linear line has the following general formula:
y = mx + c
where m is the slope and c is the y-intercept
1- getting the slope:
we are given two points which are:
(20,25) and (50,40)
the slope =
The equation now is:
y = 0.5x + c
2- getting the value of the y-intercept:
To get the value of the c, we will use any of the given points, substitute in the equation and solve for c.
I will choose the point (20,25)
y = 0.5x + c
25 = 0.5(20) + c
25 = 10 + c
c = 15
The equation of the line representing the scenario is:
y = 0.5x + 15
Now, we know that the value of the c is the y-intercept which is the initial value of the function at x=0.
In our situation, this represents the set fee.
Hope this helps :)
The set fee would be $15
Explanation:
The set fee is the starting value. This means that it is the value of the y at x = 0 (y-intercept).
To get the set fee, we would first need to get the equation of the line.
Equation of the linear line has the following general formula:
y = mx + c
where m is the slope and c is the y-intercept
1- getting the slope:
we are given two points which are:
(20,25) and (50,40)
the slope =
The equation now is:
y = 0.5x + c
2- getting the value of the y-intercept:
To get the value of the c, we will use any of the given points, substitute in the equation and solve for c.
I will choose the point (20,25)
y = 0.5x + c
25 = 0.5(20) + c
25 = 10 + c
c = 15
The equation of the line representing the scenario is:
y = 0.5x + 15
Now, we know that the value of the c is the y-intercept which is the initial value of the function at x=0.
In our situation, this represents the set fee.
Hope this helps :)