Question

Y=3x^2+30x+71

using completing the square to rewrite the given equation and reveal extreme value
y=3(x+ )^2+ .
the extreme value is at ( , )

Answer 1

(-5, -4)

Explanation:

`\sf y=3x^2+30x+71`

To rewrite the given quadratic equation in the form you mentioned and find the vertex (extreme value), we can complete the square as follows:

Start with the equation:

`\sf y=3x^2+30x+71`

First, factor out the coefficient of x² (which is 3) from the x² and x terms:

`\sf y = 3(x^2 + 10x) + 71`

Now, to complete the square, we need to add and subtract a value that will make the expression inside the parentheses a perfect square trinomial. To do that, we take half of the coefficient of x (which is 10) and square it:

`\sf \left((10)/(2)\right)^2 = 5^2 = 25`

Add and subtract 25 inside the parentheses:

`\sf y = 3(x^2 + 10x + 25 - 25) + 71`

Now, factor the perfect square trinomial inside the parentheses:

`\sf y = 3((x + 5)^2 - 25) + 71`

Distribute the 3 to both terms inside the parentheses:

`\sf y = 3(x + 5)^2 - 3(25) + 71`

Simplify the constants:

y = 3(x + 5)^2 - 75 + 71

Combine the constants:

`\sf y = 3(x + 5)^2 - 4`

Now the equation is in the desired form:

`\sf y = 3\left(x + \boxed{\sf 5}\right)^2 + \boxed{-4}`

The vertex form is
`\boxed{y = a(x - h)^2 + k}`, where (h, k) is the vertex.

In this case,

the vertex is at the point (-5, -4).

So, the extreme value (vertex) is at (-5, -4).