1. **Problem statement:** We have a quadratic function $f(x) = a \cdot x^2 + b \cdot x + c$. We know the vertex (toppunkt) is $T(2,14)$ and the tangent line at point $P(0,6)$ is $y = 8x + 6$. We need to find the coefficients $a$, $b$, and $c$. 2. **Recall the vertex form and properties:** The vertex $T(h,k)$ of a parabola $f(x) = a x^2 + b x + c$ satisfies: $$h = -\frac{b}{2a}$$ $$k = f(h) = a h^2 + b h + c$$ Also, the derivative $f'(x) = 2 a x + b$ gives the slope of the tangent at any $x$. 3. **Use vertex coordinates:** From $T(2,14)$: $$2 = -\frac{b}{2a} \implies b = -4a$$ $$14 = a \cdot 2^2 + b \cdot 2 + c = 4a + 2b + c$$ Substitute $b = -4a$: $$14 = 4a + 2(-4a) + c = 4a - 8a + c = -4a + c$$ So, $$c = 14 + 4a$$ 4. **Use point $P(0,6)$ on the parabola:** At $x=0$, $f(0) = c = 6$. From step 3, $c = 14 + 4a$, so: $$6 = 14 + 4a \implies 4a = 6 - 14 = -8 \implies a = -2$$ 5. **Find $b$ and $c$ using $a = -2$:** $$b = -4a = -4 \times (-2) = 8$$ $$c = 14 + 4a = 14 + 4 \times (-2) = 14 - 8 = 6$$ 6. **Check the slope of the tangent at $P(0,6)$:** The derivative is: $$f'(x) = 2 a x + b = 2 \times (-2) x + 8 = -4x + 8$$ At $x=0$: $$f'(0) = -4 \times 0 + 8 = 8$$ This matches the slope of the tangent line $y = 8x + 6$. **Final answer:** $$a = -2, \quad b = 8, \quad c = 6$$