What is the vertex form of #y=2x^2-18x+4 #?

Question

What is the vertex form of #y=2x^2-18x+4  #?

Kennedy Adams · Accepted Answer

#y= 2(x-9/2)^2 -73/2# #y= 2(x^2 -9x)+4# #=2 (x^2 -9x +81/4 -81/4)+4# #= 2 (x^2 -9x +81/4) -81/2 +4# #= 2(x-9/2)^2 -73/2#, which is the required form.

Nathan Axel · Answer

undefined The vertex form of a quadratic equation is given by $y = a(x-h)^2 + k$, where $(h, k)$ is the vertex of the parabola.

Given the quadratic equation $y = 2x^2 - 18x + 4$, we can convert it to vertex form by completing the square:

1. Factor out the coefficient of $x^2$, which is 2:
   $$y = 2(x^2 - 9x) + 4$$

2. To complete the square, add and subtract the square of half the coefficient of $x$, which is $\left(\frac{9}{2}\right)^2 = 20.25$, inside the parentheses. Remember to balance the equation by adding and subtracting $2 \times 20.25$, because we factored out 2 initially:
   $$y = 2(x^2 - 9x + 20.25 - 20.25) + 4$$
   $$y = 2(x^2 - 9x + 20.25) - 2 \times 20.25 + 4$$

3. Simplify the equation:
   $$y = 2(x - 4.5)^2 - 40.5 + 4$$
   $$y = 2(x - 4.5)^2 - 36.5$$

So, the vertex form of the given quadratic equation $y = 2x^2 - 18x + 4$ is $y = 2(x - 4.5)^2 - 36.5$, where the vertex $(h, k)$ of the parabola is $(4.5, -36.5)$.