Question

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.

Answer 1

(fix) "... between the parabola y = −2(x + 1)^2 - 5" **

Answer 2

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