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.
(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
To find the points of intersection between the parabola and the straight line, we need to set the two equations equal to each other and solve for the x-coordinate(s) where they intersect.
a. Let's start by setting the equations equal to each other:
-2(x + 1)^2 - 5 = mx - 7
Now simplify the equation:
-2(x^2 + 2x + 1) - 5 = mx - 7
-2x^2 - 4x - 2 - 5 = mx - 7
-2x^2 - 4x - 7 = mx - 7
-2x^2 - 4x = mx
Now rearrange the equation:
-2x^2 - (4 + m)x = 0
To find the x-coordinate(s) of the intersection point(s), we need to solve this quadratic equation. Since the equation is equal to 0, we can use the quadratic formula to solve for x:
x = (-b ± √(b^2 - 4ac)) / 2a
In this equation, a = -2, b = -(4 + m), and c = 0.
Now, substitute the values into the quadratic formula:
x = (-(4 + m) ± √((4 + m)^2 - 4(-2)(0))) / 2(-2)
x = (-(4 + m) ± √(16 + 8m + m^2)) / -4
x = (-(4 + m) ± √(m^2 + 8m + 16)) / -4
Now, we can simplify further:
x = (-4 - m ± √(m^2 + 8m + 16)) / -4
x = (-4 - m ± √((m + 4)^2)) / -4
x = (-4 - m ± (m + 4)) / -4
Simplifying even more:
x = (-4 - m + m + 4) / -4 or x = (-4 - m - m - 4) / -4
x = 0 / -4 or x = (-8 - 2m) / -4
x = 0 or x = 2 + 0.5m
Therefore, there are two possible points of intersection, (0, y) and (2 + 0.5m, y), where y is the corresponding value of y for each x-coordinate.
b. To find the value of m such that y = mx - 7 has one intersection point with y = -m(x + 1)^2 - 5, we need to set the two equations equal to each other:
mx - 7 = -m(x + 1)^2 - 5
Now simplify the equation:
mx - 7 = -m(x^2 + 2x + 1) - 5
mx - 7 = -mx^2 - 2mx - m - 5
mx + mx^2 + 2mx = -m - 5 + 7
mx^2 + (m + 2m + m)x + (m + 7 - 5) = 0
mx^2 + 4mx + (m + 2) = 0
Since we are looking for one intersection point, the quadratic equation should have only one solution. This means the discriminant (b^2 - 4ac) should be equal to zero:
(4m)^2 - 4(m)(m + 2) = 0
Now simplify and solve for m:
16m^2 - 4m^2 - 8m = 0
12m^2 - 8m = 0
4m(3m - 2) = 0
This equation will have a solution when either 4m = 0 or 3m - 2 = 0.
If 4m = 0, then m = 0.
If 3m - 2 = 0, then m = 2/3.
Therefore, m can be either 0 or 2/3.