16. a. Find how many points of intersection exist between the parabola y = −2(x + 1)2 − 5,
where y = f(x), x ∈ R, and the straight line y = mx − 7, where y = f(x), x ∈ R.
b. Find m (m < 0) such that y = mx − 7 has one intersection point with y = −m(x + 1)^ 2 − 5.
2 answers
(fix) "... between the parabola y = −2(x + 1)^2 - 5" **
a.
− 2 ( x + 1 )² − 5 = m x − 7
− 2 ( x² + 2 ∙ x ∙ 1 + 1² ) − 5 = m x − 7
− 2 ( x² + 2 x + 1 ) − 5 = m x − 7
− 2 x² - 4 x - 2 − 5 = m x − 7
− 2 x² - 4 x - 7 = m x − 7
Add 7 to both sides
− 2 x² - 4 x - 7 + 7 = m x − 7 + 7
− 2 x² - 4 x = m x
x ( - 2 x - 4 ) = m x
Divide both sides by x
- 2 x - 4 = m
m = - 2 x - 4
m = - 2 ( x + 2 )
- 2 ( x + 2 ) is defined for all x ∈ R
It has infinitely many points of intersection between
the parabola y = − 2 x² - 4 x - 2 − 5 and the straight line y = mx − 7
b.
− m ( x + 1 )² − 5 = m x − 7
Add 5 to both sides
− m ( x + 1 )² − 5 + 5 = m x − 7 + 5
− m ( x + 1 )² = m x − 2
subtract m x to both sides
− m ( x + 1 )² - m x = m x − 2 - m x
− m ( x + 1 )² - m x = − 2
− m ( x² + 2 ∙ x ∙ 1 + 1² ) - m x = − 2
− m ( x² + 2 x + 1 ) - m x = − 2
− m x² - 2 m x - m - m x = − 2
− m x² - 3 m x - m = − 2
Add 2 t o both sides
− m x² - 3 m x - m + 2 = − 2 + 2
− m x² - 3 m x - m + 2 = 0
Multiply both sides by - 1
m x² + 3 m x + m - 2 = 0
Quadratic equation a x² + b x + c = 0
have one solution if discriminant D = b² - 4 a c = 0
in this case: a = m , b = 3 m , c = m - 2
so
D = ( 3 m )² - 4 ∙ m ∙ ( m - 2 ) = 0
9 m² - 4 m ∙ m - 4 m ∙ ( - 2 ) = 0
9 m² - 4 m² + 8 m = 0
5 m² + 8 m = 0
m ∙ ( 5 m + 8 ) = 0
The solutions are:
m = 0
and
5 m + 8 = 0
5 m = - 8
m = - 8 / 5
You want solution m < 0
so m = - 8 / 5
− 2 ( x + 1 )² − 5 = m x − 7
− 2 ( x² + 2 ∙ x ∙ 1 + 1² ) − 5 = m x − 7
− 2 ( x² + 2 x + 1 ) − 5 = m x − 7
− 2 x² - 4 x - 2 − 5 = m x − 7
− 2 x² - 4 x - 7 = m x − 7
Add 7 to both sides
− 2 x² - 4 x - 7 + 7 = m x − 7 + 7
− 2 x² - 4 x = m x
x ( - 2 x - 4 ) = m x
Divide both sides by x
- 2 x - 4 = m
m = - 2 x - 4
m = - 2 ( x + 2 )
- 2 ( x + 2 ) is defined for all x ∈ R
It has infinitely many points of intersection between
the parabola y = − 2 x² - 4 x - 2 − 5 and the straight line y = mx − 7
b.
− m ( x + 1 )² − 5 = m x − 7
Add 5 to both sides
− m ( x + 1 )² − 5 + 5 = m x − 7 + 5
− m ( x + 1 )² = m x − 2
subtract m x to both sides
− m ( x + 1 )² - m x = m x − 2 - m x
− m ( x + 1 )² - m x = − 2
− m ( x² + 2 ∙ x ∙ 1 + 1² ) - m x = − 2
− m ( x² + 2 x + 1 ) - m x = − 2
− m x² - 2 m x - m - m x = − 2
− m x² - 3 m x - m = − 2
Add 2 t o both sides
− m x² - 3 m x - m + 2 = − 2 + 2
− m x² - 3 m x - m + 2 = 0
Multiply both sides by - 1
m x² + 3 m x + m - 2 = 0
Quadratic equation a x² + b x + c = 0
have one solution if discriminant D = b² - 4 a c = 0
in this case: a = m , b = 3 m , c = m - 2
so
D = ( 3 m )² - 4 ∙ m ∙ ( m - 2 ) = 0
9 m² - 4 m ∙ m - 4 m ∙ ( - 2 ) = 0
9 m² - 4 m² + 8 m = 0
5 m² + 8 m = 0
m ∙ ( 5 m + 8 ) = 0
The solutions are:
m = 0
and
5 m + 8 = 0
5 m = - 8
m = - 8 / 5
You want solution m < 0
so m = - 8 / 5