1. **Stating the problem:** We are given functions $f(x)$ and points $P(x_0, y_0)$, and we need to find the equations of the tangent lines to the graph of $f$ that pass through $P$. We also find the points of contact on the graph. 2. **Formula and rules:** The tangent line to $f$ at point $x=a$ has slope $f'(a)$ and equation: $$y = f'(a)(x - a) + f(a)$$ We want this line to pass through $P(x_0, y_0)$, so: $$y_0 = f'(a)(x_0 - a) + f(a)$$ This gives an equation in $a$ to solve for the point of contact. 3. **Apply to problem (a):** - $f(x) = -x^2 + 5$ - $P(0, 9)$ Calculate derivative: $$f'(x) = -2x$$ Equation of tangent at $x=a$: $$y = -2a(x - a) + (-a^2 + 5) = -2a x + 2a^2 - a^2 + 5 = -2a x + a^2 + 5$$ Since tangent passes through $P(0,9)$: $$9 = -2a \cdot 0 + a^2 + 5$$ $$9 = a^2 + 5$$ $$a^2 = 4$$ $$a = \pm 2$$ Points of contact: - For $a=2$: $f(2) = -4 + 5 = 1$ - For $a=-2$: $f(-2) = -4 + 5 = 1$ Tangent lines: - For $a=2$: $$y = -2 \cdot 2 (x - 2) + f(2) = -4(x - 2) + 1 = -4x + 8 + 1 = -4x + 9$$ - For $a=-2$: $$y = -2 \cdot (-2)(x + 2) + f(-2) = 4(x + 2) + 1 = 4x + 8 + 1 = 4x + 9$$ **Answer (a):** Tangent lines are $y = -4x + 9$ and $y = 4x + 9$ with points of contact $(2,1)$ and $(-2,1)$. 4. **Summary:** We found the tangent lines passing through $P$ by solving for $a$ in the tangent line equation and then writing the tangent line equations. Since the user asked multiple parts but per instructions we solve only the first problem completely.