1. **State the problem:** We have a quadratic function $$f(x) = -2x^2 + px - 10$$ and a line $$l: y = 4x + 8$$. We want to find the value(s) of $$p$$ such that the line $$l$$ is tangent to the graph of $$f$$, and find the coordinates of the tangent point(s). 2. **Recall the condition for tangency:** A line is tangent to a curve if they intersect at exactly one point. This means the quadratic equation formed by setting $$f(x) = l(x)$$ has exactly one solution. 3. **Set the equations equal:** $$-2x^2 + px - 10 = 4x + 8$$ 4. **Bring all terms to one side:** $$-2x^2 + px - 10 - 4x - 8 = 0$$ $$-2x^2 + (p - 4)x - 18 = 0$$ 5. **Use the discriminant condition for one solution:** The quadratic equation $$ax^2 + bx + c = 0$$ has one solution if its discriminant $$\Delta = b^2 - 4ac = 0$$. Here, $$a = -2$$, $$b = p - 4$$, $$c = -18$$. Calculate discriminant: $$\Delta = (p - 4)^2 - 4(-2)(-18) = (p - 4)^2 - 144$$ Set $$\Delta = 0$$: $$ (p - 4)^2 - 144 = 0 $$ 6. **Solve for $$p$$:** $$ (p - 4)^2 = 144 $$ $$ p - 4 = \pm 12 $$ So, $$ p = 4 + 12 = 16 $$ or $$ p = 4 - 12 = -8 $$ 7. **Find tangent point(s) for each $$p$$:** - For $$p = 16$$: Quadratic equation: $$ -2x^2 + (16 - 4)x - 18 = 0 $$ $$ -2x^2 + 12x - 18 = 0 $$ Divide both sides by $$-2$$: $$ \cancel{-2}x^2 + \cancel{12}x - \cancel{18} = 0 \Rightarrow x^2 - 6x + 9 = 0 $$ This factors as: $$ (x - 3)^2 = 0 $$ So, $$ x = 3 $$. Find $$y$$ using line equation: $$ y = 4(3) + 8 = 12 + 8 = 20 $$ Tangent point: $$(3, 20)$$. - For $$p = -8$$: Quadratic equation: $$ -2x^2 + (-8 - 4)x - 18 = 0 $$ $$ -2x^2 - 12x - 18 = 0 $$ Divide both sides by $$-2$$: $$ \cancel{-2}x^2 - \cancel{12}x - \cancel{18} = 0 \Rightarrow x^2 + 6x + 9 = 0 $$ This factors as: $$ (x + 3)^2 = 0 $$ So, $$ x = -3 $$. Find $$y$$ using line equation: $$ y = 4(-3) + 8 = -12 + 8 = -4 $$ Tangent point: $$(-3, -4)$$. **Final answer:** $$ p = 16 $$ with tangent point $$(3, 20)$$ $$ p = -8 $$ with tangent point $$(-3, -4)$$