1. **State the problem:** We have two functions: $$f(x) = x^2 + 3x - 5$$ $$g(x) = 2x - 3$$ We need to find: a) $f(-1)$ b) $f(0)$ c) $g\left(\frac{1}{2}\right)$ d) $f(2b)$ e) $g(1 - 4a)$ f) $x$ when $f(x) = g(x)$ 2. **Recall the formulas:** - To find $f(c)$, substitute $x=c$ into $f(x)$. - To find $g(c)$, substitute $x=c$ into $g(x)$. - To solve $f(x) = g(x)$, set the expressions equal and solve for $x$. 3. **Calculate each part:** **a) Calculate $f(-1)$:** $$f(-1) = (-1)^2 + 3(-1) - 5 = 1 - 3 - 5 = -7$$ **b) Calculate $f(0)$:** $$f(0) = 0^2 + 3(0) - 5 = 0 + 0 - 5 = -5$$ **c) Calculate $g\left(\frac{1}{2}\right)$:** $$g\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right) - 3 = 1 - 3 = -2$$ **d) Calculate $f(2b)$:** $$f(2b) = (2b)^2 + 3(2b) - 5 = 4b^2 + 6b - 5$$ **e) Calculate $g(1 - 4a)$:** $$g(1 - 4a) = 2(1 - 4a) - 3 = 2 - 8a - 3 = -8a - 1$$ **f) Solve $f(x) = g(x)$:** Set equal: $$x^2 + 3x - 5 = 2x - 3$$ Bring all terms to one side: $$x^2 + 3x - 5 - 2x + 3 = 0$$ Simplify: $$x^2 + x - 2 = 0$$ Factor: $$(x + 2)(x - 1) = 0$$ Solutions: $$x = -2 \quad \text{or} \quad x = 1$$ **Final answers:** - a) $f(-1) = -7$ - b) $f(0) = -5$ - c) $g\left(\frac{1}{2}\right) = -2$ - d) $f(2b) = 4b^2 + 6b - 5$ - e) $g(1 - 4a) = -8a - 1$ - f) $x = -2$ or $x = 1$