What is the vertex of #y=x^2-x+9-2(x-3)^2 #?

Question

What is the vertex of #y=x^2-x+9-2(x-3)^2  #?

Kennedy Adams · Accepted Answer

Vertex #(11/2, 85/4)#

Given -

#y=x^2-x+9-2(x-3)^2#

#y=x^2-x+9-2(x^2-6x+9)# #y=x^2-x+9-2x^2+12x-18# #y=-x^2+11x-9#

Vertex

#x=(-b)/(2a)=(-11)/(2 xx(-1))=11/2#

#y=-(11/2)^2+11((11)/2)-9#

#y=-121/4+121/2-9=(-121+242-36)/4=85/4#

Vertex #(11/2, 85/4)#

Eva Adamson · Answer

#(11/2, 85/4)# Simplify to #y=ax^2+bx+c# form. #y=x^2-x+9-2(x-3)^2# Use FOIL to expand #-2(x-3)^2# #y=x^2-x+9-2(x^2-6x+9)# #y=x^2-x+9-2x^2+12x-18# Combine like terms #y=-x^2+11x-9# Now that we have turned the equation to #y=ax^2+bx+c# form, Let's turn them to #y=a(x-p)^2+q# form which will give the vertex as #(p, q)#. #y=-(x^2-11x+?)-9+?# To make perfect square like #(x-p)^2#, We need to find out what #?# is. We know the formula that when #x^2-ax+b# is factorable by perfect square #(x-a/2)^2#, we get the relationship between #a# and #b#. #b=(-a/2)^2# So #b# becomes #?# and #a# becomes #-11#. Substitute those values and let's find #?#. #?=(-11/2)^2# #?=(-11)^2/(2)^2# #∴?=121/4# Substitute #?=121/4# to #y=-(x^2-11x+?)-9+?# #y=-(x^2-11x+121/4)-9+121/4# #y=-(x-11/2)^2-36/4+121/4# #y=-(x-11/2)^2+85/4# #∴y=-(x-11/2)^2+85/4# Therefore, we have turned the equation to #y=a(x-p)^2+q# form which will give our vertex as #(p, q)# #∴p=11/2, q=85/4# #∴Vertex (11/2, 85/4)#

Genesis Allen · Answer

#(5.5, 21.25)#

This equation looks scary, which makes it hard to work with. So, what we're gonna do is simplify it as far as we can and then use a small part of the quadratic formula to find the #x#-value of the vertex, and then plug that into the equation to get out our #y#-value.

Let's start with simplifying this equation:

At the end, there's this part: #-2(x-3)^2#

Which we can factor to #-2(x^2-6x+9)# (remember it isn't just #-2(x^2+9)#)

When we distribute that #-2#, we finally get out #-2x^2+12x-18#.

Put that back into the original equation and we get:

#x^2-x+9-2x^2+12x-18#, which still looks a bit scary.

However, we can simplify it down to something very recognizable:

#-x^2+11x-9# comes together when we combine all the like terms.

Now comes the cool part:

A small piece of the quadratic formula called the vertex equation can tell us the x-value of the vertex. That piece is #(-b)/(2a)#, where #b# and #a# come from the standard quadratic form #f(x)=ax^2+bx+c#.

Our #a# and #b# terms are #-1# and #11#, respectively.

We come out with #(-(11))/(2(-1))#, which comes down to

#(-11)/(-2)#, or #5.5#.

With knowing #5.5# as our vertex's #x#-value, we can plug that into our equation to get the corresponding #y#-value:

#y=-(5.5)^2+11(5.5)-9#

Which goes to:

#y=-30.25+60.5-9#

Which goes to:

#y=21.25#

Pair that with the #x#-value we just plugged in, and you get your final answer of:

#(5.5,21.25)#

Eli Assouline · Answer

undefined To find the vertex of the quadratic function $y = x^2 - x + 9 - 2(x - 3)^2$, we can rewrite the function in standard form $y = ax^2 + bx + c$, where $a$, $b$, and $c$ are constants:

First, expand the squared term $2(x - 3)^2$:
$$2(x - 3)^2 = 2(x^2 - 6x + 9) = 2x^2 - 12x + 18$$

Now, substitute this expression back into the original function:
$$y = x^2 - x + 9 - (2x^2 - 12x + 18)$$
$$y = x^2 - x + 9 - 2x^2 + 12x - 18$$
$$y = -x^2 + 11x - 9$$

Now, we have the quadratic function in standard form. To find the vertex, we can use the formula:
$$x_v = -\frac{b}{2a}$$

From the standard form, $a = -1$ and $b = 11$, so:
$$x_v = -\frac{11}{2(-1)} = \frac{11}{2}$$

Now, we can find the corresponding $y$ value by substituting $x_v$ into the equation:
$$y = -\left(\frac{11}{2}\right)^2 + 11\left(\frac{11}{2}\right) - 9$$
$$y = -\frac{121}{4} + \frac{121}{2} - 9$$
$$y = -\frac{121}{4} + \frac{242}{4} - \frac{36}{4}$$
$$y = \frac{85}{4}$$

Therefore, the vertex of the function $y = x^2 - x + 9 - 2(x - 3)^2$ is $(\frac{11}{2}, \frac{85}{4})$.