1. **State the problem:** We have a quadratic function $y=f(x)$ with axis of symmetry $x=4$, y-intercept at $(0,12)$, and passing through $(1,5)$. We need to find: a) The roots of $f(x)=0$. b) The minimum value of the function. 2. **Use the vertex form of a quadratic:** Since the axis of symmetry is $x=4$, the vertex form is: $$f(x) = a(x-4)^2 + k$$ where $(4,k)$ is the vertex. 3. **Use the y-intercept to find $a$ and $k$:** At $x=0$, $f(0)=12$: $$12 = a(0-4)^2 + k = 16a + k$$ 4. **Use the point $(1,5)$ to get another equation:** $$5 = a(1-4)^2 + k = 9a + k$$ 5. **Solve the system:** From step 3: $k = 12 - 16a$ Substitute into step 4: $$5 = 9a + 12 - 16a = 12 - 7a$$ $$5 - 12 = -7a$$ $$-7 = -7a$$ $$a = 1$$ Then, $$k = 12 - 16(1) = 12 - 16 = -4$$ 6. **Write the quadratic function:** $$f(x) = (x-4)^2 - 4$$ 7. **Find the roots by solving $f(x)=0$:** $$0 = (x-4)^2 - 4$$ $$ (x-4)^2 = 4$$ $$x-4 = \pm 2$$ $$x = 4 \pm 2$$ So, $$x=6 \text{ or } x=2$$ 8. **Find the minimum value:** Since $a=1 > 0$, the parabola opens upwards, so the vertex is a minimum. Minimum value is $f(4) = (4-4)^2 - 4 = -4$ **Final answers:** - Roots: $x=2$ and $x=6$ - Minimum value: $-4$ at $x=4$