9Number 9ExplanationLine hhh passes through point T(−1,3)T(-1,3)T(−1,3) and has a gradient mmm. For hhh to intersect the graph y=−x2y = -x^2y=−x2 at two distinct points, it must be that...m>6m > 6m>6−2<m<6-2 < m < 6−2<m<6−6<m<2-6 < m < 2−6<m<2m≤−2∨m≥6m \leq -2 \lor m \geq 6m≤−2∨m≥6m<−2∨m>6m < -2 \lor m > 6m<−2∨m>6ExplanationThe equation of line hhh is y=mx+cy = mx + cy=mx+c. Given y=3y = 3y=3 and x=−1x = -1x=−1, then: 3=−m+cc=m+3\begin{aligned} 3 &= -m + c \\ c &= m + 3 \end{aligned}3c=−m+c=m+3 Thus, the line equation can be written as y=mx+(m+3)y = mx + (m + 3)y=mx+(m+3). Since line hhh intersects the graph y=−x2y = -x^2y=−x2 at two distinct points, then: mx+(m+3)=−x2x2+mx+(m+3)=0\begin{aligned} mx + (m + 3) &= -x^2 \\ x^2 + mx + (m + 3) &= 0 \end{aligned}mx+(m+3)x2+mx+(m+3)=−x2=0 For the line to intersect at two distinct points, the discriminant must satisfy D>0D > 0D>0. m2−4(m+3)>0m2−4m−12>0(m+2)(m−6)>0\begin{aligned} m^2 - 4(m + 3) &> 0 \\ m^2 - 4m - 12 &> 0 \\ (m + 2)(m - 6) &> 0 \end{aligned}m2−4(m+3)m2−4m−12(m+2)(m−6)>0>0>0 The roots are m=−2m = -2m=−2 and m=6m = 6m=6. By testing intervals, we obtain the solution set: m<−2∨m>6m < -2 \lor m > 6m<−2∨m>6
9Number 9ExplanationLine hhh passes through point T(−1,3)T(-1,3)T(−1,3) and has a gradient mmm. For hhh to intersect the graph y=−x2y = -x^2y=−x2 at two distinct points, it must be that...m>6m > 6m>6−2<m<6-2 < m < 6−2<m<6−6<m<2-6 < m < 2−6<m<2m≤−2∨m≥6m \leq -2 \lor m \geq 6m≤−2∨m≥6m<−2∨m>6m < -2 \lor m > 6m<−2∨m>6ExplanationThe equation of line hhh is y=mx+cy = mx + cy=mx+c. Given y=3y = 3y=3 and x=−1x = -1x=−1, then: 3=−m+cc=m+3\begin{aligned} 3 &= -m + c \\ c &= m + 3 \end{aligned}3c=−m+c=m+3 Thus, the line equation can be written as y=mx+(m+3)y = mx + (m + 3)y=mx+(m+3). Since line hhh intersects the graph y=−x2y = -x^2y=−x2 at two distinct points, then: mx+(m+3)=−x2x2+mx+(m+3)=0\begin{aligned} mx + (m + 3) &= -x^2 \\ x^2 + mx + (m + 3) &= 0 \end{aligned}mx+(m+3)x2+mx+(m+3)=−x2=0 For the line to intersect at two distinct points, the discriminant must satisfy D>0D > 0D>0. m2−4(m+3)>0m2−4m−12>0(m+2)(m−6)>0\begin{aligned} m^2 - 4(m + 3) &> 0 \\ m^2 - 4m - 12 &> 0 \\ (m + 2)(m - 6) &> 0 \end{aligned}m2−4(m+3)m2−4m−12(m+2)(m−6)>0>0>0 The roots are m=−2m = -2m=−2 and m=6m = 6m=6. By testing intervals, we obtain the solution set: m<−2∨m>6m < -2 \lor m > 6m<−2∨m>6
