Convert each of the following quadratic functions to vertex form by completing the square.𝑦 =2/3 𝑥^2 − 6𝑥 + 4

Answer 1

To convert the quadratic function 𝑦 = 2/3 𝑥^2 − 6𝑥 + 4 to vertex form, we need to complete the square.

First, let's factor out the common factor of 2/3 from the first two terms:

𝑦 = (2/3)(𝑥^2 − 9𝑥) + 4

Now, we can complete the square inside the parentheses. To do this, we need to take half of the coefficient of the middle term (-9), square it, and add it inside the parentheses. Half of -9 is -9/2, and (-9/2)^2 = 81/4.

So, by adding 81/4 inside the parentheses, we would need to subtract (2/3)(81/4) outside the parentheses to maintain the equality.

𝑦 = (2/3)(𝑥^2 − 9𝑥 + 81/4 − 81/4) + 4

Simplifying inside the parentheses, we have:

𝑦 = (2/3)(𝑥^2 − 9𝑥 + 81/4) - (2/3)(81/4) + 4

Now, let's rewrite the trinomial inside the parentheses as a perfect square:

𝑦 = (2/3)(𝑥 - 9/2)^2 - (162/12) + 4

Simplifying further:

𝑦 = (2/3)(𝑥 - 9/2)^2 - 162/12 + 48/12

Finally, combining the fractions:

𝑦 = (2/3)(𝑥 - 9/2)^2 - 114/12

The quadratic function 𝑦 = 2/3 𝑥^2 − 6𝑥 + 4, in vertex form, is:

𝑦 = (2/3)(𝑥 - 9/2)^2 - 19/6