The graphs of y = x^2 + 5x + 6 and y = mx – 3, where m is a constant, are plotted on the same set of axes. Given that the graphs do not meet, what is the complete range of possible... The graphs of y = x^2 + 5x + 6 and y = mx – 3, where m is a constant, are plotted on the same set of axes. Given that the graphs do not meet, what is the complete range of possible values of m? Read more

Steps to Solve

1. Set the equations equal to each other To find the points of intersection, we set the two equations equal: $$x^2 + 5x + 6 = mx - 3$$

2. Rearrange the equation into standard quadratic form Rearrange the equation to get a quadratic equation in the standard form $ax^2 + bx + c = 0$: $$x^2 + 5x - mx + 6 + 3 = 0$$ $$x^2 + (5-m)x + 9 = 0$$

3. Apply the discriminant condition for no real roots For the quadratic equation to have no real roots (i.e., the graphs do not intersect), the discriminant ($b^2 - 4ac$) must be less than 0. In this case, $a = 1$, $b = 5-m$, and $c = 9$. $$(5-m)^2 - 4(1)(9) < 0$$

4. Expand and simplify the inequality Expand the square and simplify the inequality: $$25 - 10m + m^2 - 36 < 0$$ $$m^2 - 10m - 11 < 0$$

5. Factor the quadratic inequality Factor the quadratic expression: $$(m - 11)(m + 1) < 0$$

6. Determine the range of values for m To solve the inequality $(m - 11)(m + 1) < 0$, we analyze the sign of the expression $(m - 11)(m + 1)$. The critical points are $m = -1$ and $m = 11$.

• If $m < -1$, both $(m - 11)$ and $(m + 1)$ are negative, so their product is positive.
• If $-1 < m < 11$, $(m - 11)$ is negative and $(m + 1)$ is positive, so their product is negative.
• If $m > 11$, both $(m - 11)$ and $(m + 1)$ are positive, so their product is positive. Therefore, the inequality $(m - 11)(m + 1) < 0$ holds when $-1 < m < 11$.