Question

F(x)=4x²+6t-1
Rewrite in standard form and give the vertex

Answer 1

Assuming the intended function is f(x)=4x²+6x-1, the standard form after completing the square is f(x)=4(x+0.75)^2 - 4.75, and the vertex of the parabola is (-0.75, -4.75).

Step-by-step explanation:

To rewrite the function f(x)=4x²+6t-1 in standard form and find the vertex, we need to complete the square. However, there seems to be a typo as we usually expect the function to be in terms of one variable, either x or t, not both. Assuming the correct form is f(x)=4x²+6x-1, we can complete the square to put it in the vertex form, which is y=a(x-h)^2+k, where (h, k) is the vertex.

First, we factor out the coefficient of the x² term from the x terms:

f(x)=4(x²+⅓x)-1

Next, to complete the square, we add and subtract (⅓/2)^2 inside the parentheses:

f(x)=4((x+⅓/2)^2 - (⅓/2)^2) - 1

f(x)=4((x+⅓/2)^2 - ⅓/16) - 1

Then we simplify the equation:

f(x)=4(x+⅓/2)^2 - 4(⅓/16) - 1

f(x)=4(x+⅓/2)^2 - ⅓ - 1

f(x)=4(x+⅓/2)^2 - (⅓+16)/4

Finally, we have the standard form:

f(x)=4(x+0.75)^2 - 4.75

The vertex is at the point (-0.75, -4.75), which is the point (h, k) in the vertex form.