For the function, find the vertex, axis of symetry, and the max or minimum value.?

Question

f(x) = 2x^2 - 7x + 2

The coordinates for the vertex must be fractions.

Answer 1

If you differentiate (I don't know if you know calc) you get
4 x-7.

If you set this equal to zero, you get the minumum & the vertex

4x-7=0

x=7/4 (this is where the vertex & the minimum are)

Plug this into f(x) to get the y value:

F(7/4)= 2 (7/4)^2 - 7(7/4)+2= -33/8
So the answers are :

Min at (7/4, -33/8) = Vertex

Axis of sym x= 7/4
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If you don't know calc, you get the same answer by finding the 2 roots.

Then, the minimum, axis of symmetry, & vertex occur halfway in-between the 2 roots

Answer 2

f(x) = 2x^2 - 7x + 2

To find the min/max of this function without using Calculus, you have to use the method of completing the square.

Your first step is to factor the coefficient of x^2 from the first two terms.

f(x) = 2(x^2 - (7/2)x) + 2

Now, add and subtract half squared of the coefficient of x. In our case, the coefficient of x is -7, so half would be (-7/2), square would be 49/4. Add and subtract this amount.

f(x) = 2(x^2 - (7/2)x + 49/4 - 49/4) + 2

Now that we have a complete square, we factor it as so.

f(x) = 2( (x - 7/4)^2 - 49/4 ) + 2

Distribute the 2.

f(x) = 2(x - 7/4)^2 - 2(49/4) + 2

f(x) = 2(x - 7/4)^2 - 49/2 + 2

f(x) = 2(x - 7/4)^2 - 49/2 + 4/2

f(x) = 2(x - 7/4)^2 - 45/2

For a parabola of the form f(x) + a(x - h)^2 + k, the vertex is located at (h, k). If a > 0, then (h, k) is a minimum; if a < 0, then (h, k) is a maximum. The minimum/maximum value is k.

Since a = 2, a is positive, so we have a minimum value of -45/2.

Answer 3

Start with the axis of symmetry

x = -b/2a

In our question
a=2
b=-7
c=2

x=-(-7)/(2(2)) = 7/4

This is the axis of symmetry and the x coordinate of the vertex. Plug into original equation to find y coordinate.

y=2(7/4)^2 - 7(7/4) +2
y= 2(49/16) -49/4 + 2

LCD = 16
y = 98/16 - 196/16 +32/16
y = -66/16
y = -33/8

Vertex is (7/4,-33/8)

Since a is positive the parabola opens up like U so the vertex is the minimum value, and there is no maximum.

Answer 4

First, you should find out about your quadratic. Since the x^2 term is positive, it will open "up". Since b^2-4*a*c = 33, there are two roots, and since sqrt(33) is not a perfect square, both are irrational. There is a minimum value for which x=7/4 is one coordinate and f(x) for x=7/4 is the other. The axis of symmetry should be the line x=7/4. I don't know about vertex stuff.

Answer 5

This is a parabola and the vertex has coordinates (h, k) where
h = -b/(2a)
and k = f(h)
ax^2+bx+c is where you find a,b, and c
a = 2
b = -7
c = 2

h = - (-7)/(2*2) = 7/4
k = f(7/4) = 2(7/4)^2 - 7(7/4) + 2 = -33/8

Vertex at (7/4, -33/8)
Axis of symmetry runs through the vertex, x = 7/4.
The minimum/maximum occurs at the vertex.
Since A is positive (+2) the vertex is at the bottom most part of the graph, so it is a minimum and its value is -33/8.

Hope this helps!
Checkout my College Algebra DVD at www.j-digital.net

Answer 6

JDigital knows what he is talking about. Check out his CD.

f(x) is the y value for a graph of points as x,y
If you put three or four different numbers for x into the equation and plot the points on the graph you will get an idea of what the graph looks like.

It is a second degree equation which usually means there are two values for x which result in y = 0.

Using the quadratic equation is a method that always works, but is some times a bit tedious. The standard form is written as ax^2 + by + c for plugging into the quadratic equation, so for your problem a = 2, b = -7, and c = 2.

Typing in this format doesn't really allow for me to show you what the equation looks like but you can plug the numbers into the equation. There will be two answers.