How do I use the vertex formula to determine the vertex of the graph of the function and write it in standard form for #2x^2+6x-1#?

Question

Savannah Anderson · Accepted Answer

#y = 2x^2+6x-1# is already in standard form #color(white)("XXXXX")#General standard form for a parabola is #y=ax^2+bx+c# Vertex form for a parabola is #y =m(x-a)+b#
#color(white)("XXXXX")#Note: the #a and b# are not the same as in the standard form but they are constants. #y=2x^2+6x+1#
#color(white)("XXXXX")#extract #m#
#y= 2(x^2+3x) + 1#
#color(white)("XXXXX")#complete the square
#y=2(x^2+3x+(3/2)^2) +1 - 9/2# #y = 2(x+3/2)^2 -7/2# #y = 2(x-(-3))^2+(-7/2)#
#color(white)("XXXXX")#which is in vertex form with a vertex at #(-3,-7/2)#

Eva Adamson · Answer

undefined To determine the vertex of the graph of the function $f(x) = 2x^2 + 6x - 1$ using the vertex formula, follow these steps:

1. Identify the coefficients $a$, $b$, and $c$ in the quadratic equation $ax^2 + bx + c$.
2. Use the formula $x = \frac{-b}{2a}$ to find the x-coordinate of the vertex.
3. Substitute the x-coordinate obtained in step 2 into the original function to find the corresponding y-coordinate.
4. Write the vertex in standard form $(h, k)$, where $h$ is the x-coordinate and $k$ is the y-coordinate.

Using these steps, the x-coordinate of the vertex is $x = \frac{-6}{2(2)} = -\frac{3}{2}$.
Substituting $x = -\frac{3}{2}$ into the original function:
$$f\left(-\frac{3}{2}\right) = 2\left(-\frac{3}{2}\right)^2 + 6\left(-\frac{3}{2}\right) - 1 = -5$$
So, the vertex is $\left(-\frac{3}{2}, -5\right)$ in standard form.