For a linear function, the rate of change (slope) is constant. Use the two points to find how much the cost increases per hour.

Once you know the cost per hour, decide whether it’s easier to: (1) build an equation for $A(t)$, or (2) start from one of the known costs and add the cost of extra hours to reach 8 hours.

In a new expression line, type (195-90)/(5-2) and note the value Desmos gives. This is the constant hourly rate (slope).

In another line, type A(t) = 90 + ((195-90)/(5-2))*(t-2). This defines the linear function that goes through the point (2, 90) with the correct slope.

In a new line, type A(8). The numerical result that appears is the total amount the tutor charges for 8 hours of tutoring.

The amount the tutor charges, $A(t)$, is a linear function of the number of hours $t$.

The two pieces of information given are:

• 2 hours costs $90 → point $(2, 90)$
• 5 hours costs $195 → point $(5, 195)$

We are asked to find the cost for 8 hours, which will be the value of $A(8)$ on this same line.

Because the relationship is linear, the hourly rate is constant. This constant rate is the slope of the line through the two points.

Compute the slope:

So the tutor charges $35 per additional hour of tutoring.

There are two efficient ways to move from the known information to 8 hours.

Method 1: From 5 hours to 8 hours

• From 5 hours to 8 hours is 3 more hours.
• Each extra hour costs $35.
• Extra cost:

Add this to the 5-hour cost of $195:

Method 2: Write the linear equation

• A linear function with slope 35 has the form $A(t) = 35t + b$.
• Use point $(2, 90)$ to find $b$:

So $A(t) = 35t + 20$. Then compute $A(8)$:

Both methods should give the same final total.

Finish either method:

• Method 1:

• Method 2:

So, the tutor will charge $\boxed{300}$ dollars for 8 hours of tutoring.