Respuesta :

Part A

The x intercepts always occur when y = 0.

The two x intercepts are at the locations (-2, 0) and (2, 0)

We can shorten that to p = -2 and q = 2

The intercept form of a quadratic is y = a(x-p)(x-q)

So we have y = a(x-(-2))(x-2) which turns into y = a(x+2)(x-2)

Plug in x = 0 and y = -4 from the y intercept and solve for 'a'

y = a(x+2)(x-2)

-4 = a(0+2)(0-2)

-4 = a(2)(-2)

-4 = -4a

a = -4/(-4)

a = 1

We go from y = a(x+2)(x-2) to y = (x+2)(x-2) which is one possible answer.

Let's expand that out using the FOIL rule

y = (x+2)(x-2)

y = x*x - 2*x + 2*x + 2(-2)

y = x^2 - 2x + 2x - 4

y = x^2 - 4 is another possible answer

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The two possible answers are:

y = (x+2)(x-2)

y = x^2 - 4

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Part B

We have p = -7 and q = -1 as the x intercepts. They occur when y = 0.

y = a(x-p)(x-q)

y = a(x-(-7))(x-(-1))

y = a(x+7)(x+1)

Plug in x = 0 and y = 7. Solve for 'a'

y = a(x+7)(x+1)

7 = a(0+7)(0+1)

7 = a(7)(1)

7 = 7a

a = 7/7

a = 1

We go from y = a(x+7)(x+1) to  y = (x+7)(x+1)

Expanding that out gets us

y = (x+7)(x+1)

y = x*x + 1x + 7x + 7*1

y = x^2 + 8x + 7

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The two possible answers are:

y = (x+7)(x+1)

y = x^2 + 8x + 7

Answer:

a.

• In standard form : x² - 4 = 0

• In factored form : (x - 2)(x + 2) = 0

b.

• In factored form : (x + 7)(x + 1) = 0

• In standard form :  x² + 8x + 7 = 0

Step-by-step explanation:

a)  Consider the quadratic expression:

P(x) = ax² + bx + c

• (0 , -4) ∈ Table ⇒ P(0) = -4 ⇒ a(0)² + b(0) + c = -4

⇒ c = -4

• (-1 , -3) ∈ Table ⇒ P(-1) = -3 ⇒ a(-1)² + b(-1) - 4 = -3

⇒ a - b = 1

• (1 , -3) ∈ Table ⇒ P(1) = -3 ⇒ a(1)² + b(1) - 4 = -3

⇒ a + b = 1

Solving the system for a and b :

a - b = 1

a + b = 1

⇒ 2a = 2 and 2b = 0

⇒ a = 1 and b = 0

Conclusion:

The quadratic equation P = 0 for the table is :

• In standard form : x² - 4 = 0

• In factored form : (x - 2)(x + 2) = 0

………………………………………………………………………

b)  Consider the quadratic expression:

Q(x) = a'x² + b'x + c'

Factored form :

Q(x) can be written in the form Q(x) = m(x - p)(x + q)

• (-7 , 0) ∈ Table ⇒ Q(x) = m(x − -7)(x - q) = m(x + 7)(x - q)

• (-1 , 0) ∈ Table ⇒ Q(x) = m(x + 7)(x − -1) = m(x + 7)(x + 1)

• (0 , 7) ∈ Table ⇒ Q(0) = 7 ⇒ m(0 + 7)(0 + 1) = 7

⇒ 7m = 7 ⇒ m = 1

Therefore ,

Q(x) = (x + 7)(x + 1)

Equation of the table :  (x + 7)(x + 1) = 0

Q in Standard form :

Just factor  (x + 7)(x + 1)

⇒ x² + x + 7x + 7 = x² + 8x + 7

Equation of the table :  x² + 8x + 7 = 0

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