1. **State the problem:** We have an exponential function $f(x) = ab^x$ passing through points $A(\frac{1}{4}, 3)$ and $B(\frac{1}{2}, 4.5)$. We want to find the $y$-coordinate of point $C$ where $x = \frac{7}{4}$. 2. **Formula and rules:** The general form of an exponential function is $$f(x) = ab^x$$ where $a$ is the initial value and $b$ is the base (growth factor). We use the given points to find $a$ and $b$. 3. **Use point A:** Substitute $x=\frac{1}{4}$ and $f(x)=3$: $$3 = ab^{\frac{1}{4}}$$ 4. **Use point B:** Substitute $x=\frac{1}{2}$ and $f(x)=4.5$: $$4.5 = ab^{\frac{1}{2}}$$ 5. **Divide the two equations to eliminate $a$:** $$\frac{4.5}{3} = \frac{ab^{\frac{1}{2}}}{ab^{\frac{1}{4}}} = b^{\frac{1}{2} - \frac{1}{4}} = b^{\frac{1}{4}}$$ Simplify the left side: $$\frac{4.5}{3} = 1.5$$ So, $$b^{\frac{1}{4}} = 1.5$$ 6. **Solve for $b$ by raising both sides to the power 4:** $$b = (1.5)^4 = 5.0625$$ 7. **Find $a$ using point A:** $$3 = a \times 5.0625^{\frac{1}{4}}$$ Calculate $5.0625^{\frac{1}{4}}$: $$5.0625^{\frac{1}{4}} = 1.5$$ So, $$3 = a \times 1.5$$ Divide both sides by 1.5: $$a = \frac{3}{1.5} = 2$$ 8. **Write the function:** $$f(x) = 2 \times 5.0625^x$$ 9. **Find $y$ at $x=\frac{7}{4}$:** $$f\left(\frac{7}{4}\right) = 2 \times 5.0625^{\frac{7}{4}}$$ Calculate the exponent: $$5.0625^{\frac{7}{4}} = \left(5.0625^{\frac{1}{4}}\right)^7 = 1.5^7 = 17.0859$$ Multiply by 2: $$f\left(\frac{7}{4}\right) = 2 \times 17.0859 = 34.1718$$ 10. **Round to the nearest tenth:** $$34.2$$ **Final answer:** The $y$-coordinate of point $C$ is approximately **34.2**.