How do you write #f(x)= -2x^2+20x-49# in vertex form?

Question

How do you write #f(x)= -2x^2+20x-49# in  vertex form?

Ethan Adkins · Accepted Answer

#y=-2(x-5)^2+1#

Standard Form equation: #y=ax^2+bx+c# Vertex Form equation: #y=a(x-h)^2+k#

Where #a# is equal to the #a# value of the standard form equation and #(h,k)# is equal to the vertex of the equation.

In order to convert it, let's first fill in what we know. The #a# value of the given equation is -2.

So, we have:

#y=# -2 #(x-h^2)#+#k#

In order to find the vertex, you must use the equation #-b/(2a)#

Looking at the standard form equation, #b=20# and #a=-2# So plugging in, you get #-20/(2*-2)#

Once solved, you're x value of your vertex is 5. Now, you plug 5 into your original standard form equation.

#y=-2(5)^2+20(5)-49#

Finally, your vertex is:

#(5, 1)#

Finally, plug it into your equation. #y=-2(x-5)^2+1#

Eva Adamson · Answer

undefined To write $ f(x) = -2x^2 + 20x - 49 $ in vertex form, complete the square. The vertex form of a quadratic function is $ f(x) = a(x - h)^2 + k $, where $ (h, k) $ is the vertex of the parabola.

First, factor out the common factor $ -2 $ from the quadratic terms:
$$ f(x) = -2(x^2 - 10x) - 49 $$

Complete the square inside the parentheses:
$$ f(x) = -2(x^2 - 10x + 25 - 25) - 49 $$

Simplify inside the parentheses and distribute:
$$ f(x) = -2[(x - 5)^2 - 25] - 49 $$

Apply the distributive property:
$$ f(x) = -2(x - 5)^2 + 50 - 49 $$

Combine like terms:
$$ f(x) = -2(x - 5)^2 + 1 $$

So, the vertex form of the function $ f(x) = -2x^2 + 20x - 49 $ is $ f(x) = -2(x - 5)^2 + 1 $.