Find the vertex of the parabola y=3x^2+x+5
which method did they teach you?
Completing the square or using a formula?
y=3x^2+x+5= 3(x^2 + x/3 + ? ) + 5
= 3(x^2 + x/3 + 1/36 ) + 5 -1/36
= 3(x+ 1/6)^2 + 4 35/36
vertex at x=-1/6
The method used here is completing the square. To find the vertex of the parabola y=3x^2+x+5, follow these steps:
1. Start with the equation: y=3x^2+x+5.
2. Rewrite the equation to create a perfect square trinomial in the form (x + h)^2:
y = 3(x^2 + x/3) + 5.
3. Take half of the coefficient of x (1/3 in this case) and square it to get (1/6)^2 = 1/36.
4. Add the square term inside the parentheses:
y = 3(x^2 + x/3 + 1/36) + 5.
5. To maintain the equality of the equation, subtract the added term outside the parentheses:
y = 3(x^2 + x/3 + 1/36) + 5 - 1/36.
6. Simplify the equation:
y = 3(x+ 1/6)^2 + 4 35/36.
7. From the equation, we can see that the vertex of the parabola is at the point (-1/6, 4 35/36). Therefore, the vertex of the parabola is at x = -1/6.