Let f(x) = begin{cases} -x^2 & text{for } x

Question

(B) Let the tangent line be $y = mx + c$.For $x \ge 0$,$f(x) = x^2 + 8$. Setting $mx + c = x^2 + 8$,we get $x^2 - mx + (8 - c) = 0$. For tangency,the

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$-1$

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(B) Let the tangent line be $y = mx + c$.For $x \ge 0$,$f(x) = x^2 + 8$. Setting $mx + c = x^2 + 8$,we get $x^2 - mx + (8 - c) = 0$. For tangency,the discriminant $D = 0$,so $m^2 - 4(8 - c) = 0$,which implies $m^2 = 32 - 4c$ (Equation $1$).For $x Equating the two expressions for $m^2$: $4c = 32 - 4c$,which gives $8c = 32$,so $c = 4$.Substituting $c = 4$ into Equation $2$,$m^2 = 16$,so $m = \pm 4$. Since the tangent must touch both branches,we consider the line $y = 4x + 4$.To find the $x$-intercept,set $y = 0$: $0 = 4x + 4$,which gives $x = -1$.

Let $f(x) = \begin{cases} -x^2 & \text{for } x < 0 \\ x^2 + 8 & \text{for } x \ge 0 \end{cases}$. Then the $x$-intercept of the line that is tangent to the graph of $f(x)$ is

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