How to find the x-intercepts of a parabola from the vertex, (-1,-108) and the y intercept (0,-105)?

* The x-intercepts are -7 and 5

Step-by-step explanation:

* At first lets revise the standard and general forms of the

 quadratic function which represented graphically by the parabola

- f(x) = a(x - h)² + k ⇒ standard form

- Where point (h , k) is the vertex of the parabola

- f(x) = ax² + bx + c ⇒ general form

- Where a, b, c are constant

- c is the y-intercept ⇒ means x = 0

- h = -b/2a

- k = f(h)

* Lets solve the problem

- We will find the equation of the parabola

∵ The vertex is (-1 , -108)

∴ h = -1 and k = -108

∵ y-intercept = -105

- Equate the two forms

∵ ax² + bx + c = a(x - h)² + k ⇒ solve the (   )²

∴ ax² + bx + c = a(x² - 2hx + h²) + k ⇒ open the bracket

∴ ax² + bx + c = ax² - 2ahx + ah² + k ⇒ by comparing the two sides

∴ c = ah² + k

- Substitute the value of c , h , k in it

∴ -105 = a(-1)² + -108

∴ -105 = a - 108 ⇒ add 108 to the both sides

∴ 3 = a

- Lets write the equation in the standard form

∴ y = 3(x - -1)² + -108

∴ y = 3(x + 1)² - 108

* To find the x-intercepts means the parabola intersects the x-axis,

  then put y = 0

∴ 3(x + 1)² - 108 = 0 ⇒ Add 108 to the both sides

∴ 3(x + 1)² = 108 ⇒ divide the both sides by 3

∴ (x + 1)² = 36 ⇒ take square root for both sides

∴ (x + 1) = ± 6

# x + 1 = 6  OR x + 1 = -6

∵ x + 1 = 6 ⇒ subtract 1 from both sides

∴ x = 5

∵ x + 1 = -6 ⇒ subtract 1 from both sides

∴ x = -7

* The x-intercepts are -7 and 5

∴

step-by-step explanation:

d

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}