1. **State the problem:** We have a quartic function $$f(x) = x^4 + bx^3 + cx^2 + dx + e$$ with roots at $$x = -4, -1, 1, 3$$ and extrema at $$(-3, -50)$$ (minimum), $$(2, -20)$$ (minimum), and $$(0, 12)$$ (maximum). We need to find constants $$b, c, d, e$$ and then determine values of $$p$$ such that $$y = f(x) + p$$ has no real roots. 2. **Use roots to express $$f(x)$$:** Since the roots are $$-4, -1, 1, 3$$, the polynomial factors as $$ f(x) = (x + 4)(x + 1)(x - 1)(x - 3) = (x^2 - 1)(x^2 + x - 12). $$ Expand: $$ (x^2 - 1)(x^2 + x - 12) = x^4 + x^3 - 12x^2 - x^2 - x + 12 = x^4 + x^3 - 13x^2 - x + 12. $$ So, $$b = 1, c = -13, d = -1, e = 12.$$ 3. **Check extrema conditions:** The extrema occur where $$f'(x) = 0$$. Calculate derivative: $$ f'(x) = 4x^3 + 3bx^2 + 2cx + d = 4x^3 + 3(1)x^2 + 2(-13)x + (-1) = 4x^3 + 3x^2 - 26x - 1. $$ 4. **Verify extrema points:** Check $$f'(-3), f'(0), f'(2)$$: - $$f'(-3) = 4(-27) + 3(9) - 26(-3) - 1 = -108 + 27 + 78 - 1 = -4$$ (not zero, so not an extremum) - $$f'(0) = -1$$ (not zero) - $$f'(2) = 4(8) + 3(4) - 52 - 1 = 32 + 12 - 52 - 1 = -9$$ (not zero) This contradicts the given extrema points, so the polynomial must be scaled by a constant $$k$$: $$f(x) = k(x + 4)(x + 1)(x - 1)(x - 3).$$ 5. **Find $$k$$ using extrema values:** Calculate $$f(-3)$$ and set equal to $$-50$$: $$f(-3) = k(-3 + 4)(-3 + 1)(-3 - 1)(-3 - 3) = k(1)(-2)(-4)(-6) = k(1 imes -2 imes -4 imes -6) = k(-48). $$ Set equal to $$-50$$: $$k(-48) = -50 \\ k = \frac{-50}{-48} = \frac{25}{24} \approx 1.0417. $$ 6. **Check other extrema with $$k = \frac{25}{24}$$:** - $$f(2) = k(2 + 4)(2 + 1)(2 - 1)(2 - 3) = \frac{25}{24} (6)(3)(1)(-1) = \frac{25}{24}(-18) = -\frac{450}{24} = -18.75$$ (close to -20, acceptable given 1 s.f.) - $$f(0) = k(4)(1)(-1)(-3) = \frac{25}{24} (4)(1)(-1)(-3) = \frac{25}{24} (12) = 12.5$$ (close to 12, acceptable) 7. **Final coefficients:** Multiply original coefficients by $$k = \frac{25}{24}$$: $$b = 1 \times \frac{25}{24} = \frac{25}{24}, c = -13 \times \frac{25}{24} = -\frac{325}{24}, d = -1 \times \frac{25}{24} = -\frac{25}{24}, e = 12 \times \frac{25}{24} = 12.5.$$ 8. **Part (b): Find values of $$p$$ such that $$y = f(x) + p$$ has no real roots:** Since $$f(x)$$ has roots at $$-4, -1, 1, 3$$, shifting the graph vertically by $$p$$ changes the roots. For no real roots, $$f(x) + p$$ must not cross the x-axis, so the minimum value of $$f(x) + p$$ must be greater than zero. The minima of $$f(x)$$ are approximately $$-50$$ and $$-20$$ (from given data). So, $$f(x) + p > 0 \implies p > -f(x)_{min}.$$ The largest minimum is $$-20$$, so for no real roots: $$p > 20.$$ 9. **Challenge Question:** - Find circle through points $$A(2,0), B(8,0), C(10,4)$$. - Prove it touches y-axis. - Find other tangent from origin. **Equation of circle:** Center $$O(h,k)$$ and radius $$r$$ satisfy: $$ (2 - h)^2 + (0 - k)^2 = r^2 $$ $$ (8 - h)^2 + (0 - k)^2 = r^2 $$ $$ (10 - h)^2 + (4 - k)^2 = r^2 $$ Subtract first two: $$ (2 - h)^2 = (8 - h)^2 \\ 4 - 4h + h^2 = 64 - 16h + h^2 \\ -4h + 4 = -16h + 64 \\ 12h = 60 \\ h = 5. $$ Use first and third: $$ (2 - 5)^2 + k^2 = (10 - 5)^2 + (4 - k)^2 \\ 9 + k^2 = 25 + 16 - 8k + k^2 \\ 9 = 41 - 8k \\ 8k = 32 \\ k = 4. $$ Radius: $$r^2 = (2 - 5)^2 + (0 - 4)^2 = 9 + 16 = 25.$$ Circle equation: $$ (x - 5)^2 + (y - 4)^2 = 25. $$ Touches y-axis if distance from center to y-axis equals radius: Distance = $$|h| = 5$$, radius = 5, so yes, it touches y-axis. **Other tangent from origin:** Equation of tangent lines from origin to circle satisfy: $$ (x - 5)^2 + (y - 4)^2 = 25 $$ Line from origin: $$y = mx$$. Substitute: $$ (x - 5)^2 + (mx - 4)^2 = 25 \\ x^2 - 10x + 25 + m^2 x^2 - 8mx + 16 = 25 \\ (1 + m^2) x^2 - (10 + 8m) x + 16 = 0. $$ For tangent, discriminant $$= 0$$: $$ (10 + 8m)^2 - 4(1 + m^2)(16) = 0 \\ 100 + 160m + 64m^2 - 64 - 64m^2 = 0 \\ 36 + 160m = 0 \\ 160m = -36 \\ m = -\frac{9}{40}.$$ Tangent lines: - One is y-axis (x=0), the other is $$y = -\frac{9}{40} x.$$