Tangents & Curves

"tangent with the line $y=10$ " implies $f(x)$ has extreme value 10.

$\begin{aligned} f(x) &= -\dfrac {x^2}4 - 3x + 1 \\ &= -\dfrac 14(x+6)^2 + 10 \\ \implies Q &= (-6, 10) \end{aligned}$

A few points to note here:

• The value of the ordinate has been given: $(y = \purple{10})$
• Since the equation of the tangent is $y = 10$ , that means the slope of the tangent is $\red{0}$ .

To find the value of the abscissa, we must find the derivative of the function and solve for x.

$f(x) = \frac{-1}{4} x^{2} - 3x + 1$ $\frac{dy}{dx} = \frac{-1}{2}x - 3$

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$\frac{-1}{2}x - 3 = \red{0}$ $x = \purple{-6}$

$\therefore$ the coordinates are $(\purple{-6}, \purple{10})$ . Hence, the answer is $\boxed{4}$