Question

PLEASE ANSWER ASAP thanks!​

Answer 1

`\textsf{2)} \quad f(x)=a\left(x+(b)/(2a)\right)^2+c-(b^2)/(4a)`

`\textsf{3)} \quad f(x)=\left(x-2\right)^2-14`

`\textsf{4)} \quad y = 3x^2-12x+ 16`

Explanation:

Question 2

To rewrite a quadratic function given in standard form f(x) = ax² + bx + c in vertex form f(x) = a(x - h)² + k, we need to complete the square.

First, factor out the leading coefficient "a" from the first two terms:

`f(x)=a\left(x^2+(b)/(a)x\right)+c`

Now, add the square of half the coefficient of the x-term inside the parentheses, and subtract the distributed value of this outside the parentheses:

`f(x)=a\left(x^2+(b)/(a)x+\left(((b)/(a))/(2)\right)^2\right)+c-a\left(((b)/(a))/(2)\right)^2`

Simplify:

`f(x)=a\left(x^2+(b)/(a)x+\left((b)/(2a)\right)^2\right)+c-a\left((b)/(2a)\right)^2`

`f(x)=a\left(x^2+(b)/(a)x+\left((b)/(2a)\right)^2\right)+c-(b^2)/(4a)`

Factor the perfect square trinomial inside the parentheses:

`f(x)=a\left(x+(b)/(2a)\right)^2+c-(b^2)/(4a)`

Therefore, f(x) = ax² + bx + c in the form f(x) = a(x - h)² + k is:

`\large\boxed{\boxed{f(x)=a\left(x+(b)/(2a)\right)^2+c-(b^2)/(4a)}}`

`\hrulefill`

Question 3

To rewrite a quadratic function in standard form f(x) = x² - 4x - 10 in vertex form f(x) = a(x - h)² + k, we need to complete the square.

Since the leading coefficient is 1, we don't need to factor this out from the first two terms, but we can add parentheses around the first two terms for ease of explanation:

`f(x)=(x^2-4x)-10`

Now, add the square of half the coefficient of the x-term inside the parentheses, and subtract the same value outside the parentheses:

`f(x)=\left(x^2-4x+\left((-4)/(2)\right)^2\right)-10-\left((-4)/(2)\right)^2`

`f(x)=\left(x^2-4x+\left(-2\right)^2\right)-10-\left(-2\right)^2`

`f(x)=\left(x^2-4x+4\right)-10-4`

`f(x)=\left(x^2-4x+4\right)-14`

Factor the perfect square trinomial inside the parentheses:

`f(x)=\left(x-2\right)^2-14`

Therefore, f(x) = x² - 4x - 10 in the form f(x) = a(x - h)² + k is:

`\large\boxed{\boxed{f(x)=\left(x-2\right)^2-14}}`

`\hrulefill`

Question 4

To rewrite y = 3(x - 2)² + 4 in the general form y = ax² + bx + c, we can begin by expanding the brackets:

`y = 3(x - 2)^2 + 4`

`y = 3(x - 2)(x-2) + 4`

`y = 3(x^2-2x-2x+4) + 4`

`y = 3(x^2-4x+4) + 4`

Distribute the 3 to the terms inside the brackets:

`y = 3\cdot x^2+3\cdot (-4x)+ 3\cdot 4 + 4`

`y = 3x^2-12x+ 12+ 4`

Simplify:

`y = 3x^2-12x+ 16`

Therefore, y = 3(x - 2)² + 4 in the general form y = ax² + bx + c is:

`\large\boxed{\boxed{y = 3x^2-12x+ 16}}`