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