The vertex of this parabola is at (2,-1) when the y value is 0 and then x value is 5 what is the coefficient of the squared term in the parabolas equation

now, let's expand the squared term to get the standard form of the quadratic.

Step-by-step explanation:

Remark

The general formula for the vertex of a parabola is

y = a(x - b) + c

What you know from the vertex is the formula describes b and c and the y intercept gives you the information to solve for a.

Solution

y = a(x - 2)^2 - 1 If you graph this, the vertex should be at (2,-1). I'll upload the graph at the end.

Now we need to solve for a

x = 0

y = 5

5 = a(0 - 2)^2 - 1

5 = a(4) - 1              Add 1 to both sides.

6 = a(4)                 Divide by 4

6/4 = a 4/4           Switch

a = 3/2

Answer

y = 3/2 (x - 2)^2 - 1

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.