Let $B(x)=1.5^x+9-2x$. The intersections are the solutions to $B(x)=b$. How many solutions happen when $b$ is below, at, or above the minimum of $B(x)$?

Graph $y=1.5^x+9-2x$ (or use a table). © аnikο⁠.аi Find the smallest $y$-value it reaches, then pick the smallest integer strictly greater than that value.

Enter

• $y_1=1.5^x+9$
• $y_2=2x+b$

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When you type $b$, Desmos will create a slider for it.

Move the slider for $b$ downward until the graphs stop intersecting. Then move it upward slowly to see the smallest $b$ values where two intersection points appear.

Since $b$ must be an integer, test $b=6$, $b=7$, and $b=8$ on the slider and count the intersection points each time.

Instead enter $y_3=1.5^x+9-2x$. The minimum $y$-value of this graph is the cutoff. Choose the smallest integer greater than that minimum so that $y_3=b$ would intersect twice.

At intersection points, $f(x)=g(x)$:

Solve for $b$:

So each $x$ gives the value of $b$ that makes the line pass through the point $(x,f(x))$.

Consider the function

For a given integer $b$, intersections correspond to solutions of $B(x)=b$.

aniкo.‌ai

Because $1.5^x$ bends upward while $-2x$ is linear, $B(x)$ decreases at first and then increases, so it has a single lowest point (a minimum). That means:

• If $b$ is below the minimum of $B(x)$, there are no intersections.
• If $b$ equals the minimum, there is one intersection (the graphs just touch).
• If $b$ is above the minimum, there are two intersections.

Use a calculator/Desmos to graph $y=B(x)=1.5^x+9-2x$ and estimate its lowest value.

From the graph (or a table of values near where it bottoms out), the minimum occurs near $x\approx 3.94$ and the minimum value is approximately

To get two intersections, $b$ must be greater than about $6.06$.

The least integer greater than $6.06$ is $7$.

Therefore, the least possible value of $b$ is $7$.