Exercise
Given the parabolic equation:
y=(m+6)x^2-(3m+2)x+2
For which values of parameter m, the parabola is on the x-axis or above it.
Final Answer
Solution
Given the parabolic equation:
y=(m+6)x^2-(3m+2)x+2
We want to find out when:
y\geq 0
Or
(m+6)x^2-(3m+2)x+2\geq 0
Two conditions are required:
1. The parabola is “smiling” – the coefficient of the squared expression is greater than zero:
m+6>0
m>-6
2. The quadratic equation do not have a solution, meaning:
\Delta<0
Because in such a situation the whole parabola is always above the x-axis.
The coefficients of the quadratic equation are
a=m+6, b=3m+2, c=2
Putting the coefficients in the Delta formula (in the quadratic formula) gives us
\Delta=b^2-4ac=
={(3m+2)}^2-4\cdot (m+6)\cdot 2=
=9m^2+12m+4-8m-48=
=9m^2+4m-44
And we want it to take place:
9m^2+4m-44<0
It is a square inequality. Its coefficients are:
a=9, b=4, c=-44
The coefficient of the squared expression (a) is positive, so the parabola (quadratic equation graph) “smiles” (= bowl-shaped). The sign of the inequality means we are looking for the sections the parabola is below the x-axis. We find the solutions (= zeros = roots) of the quadratic equation using the quadratic formula. Putting the coefficients in the formula gives us
m_{1,2}=\frac{-4\pm \sqrt{4^2-4\cdot 9\cdot (-44)}}{2\cdot 9}=
=\frac{-4\pm \sqrt{1600}}{18}=
=\frac{-4\pm 40}{18}
Hence, we get the solutions:
x_1=\frac{-4+ 40}{18}=2
x_2=\frac{-4- 40}{18}=\frac{-44}{18}=-2\frac{4}{9}
Since the parabola “smiles” and we are interested in the sections below the x-axis, we get
-2\frac{4}{9}\leq m\leq 2
We combine (“and”) both results (of both conditions) and we still get that the final answer is
-2\frac{4}{9}\leq m\leq 2
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