How do you find the vertex of #f(x) = x^2 + 3#?

Question

Ethan Adkins · Accepted Answer

#(0, 3)#

The equation of a parabola (with vertical axis) in vertex form can be written:

#f(x) = a(x-h)^2 + k#

where #(h, k)# is the vertex and #a != 0# a constant multiplier.

In our case we find:

#f(x) = 1(x-0)^2+3#

which means that #(h, k) = (0, 3)# and #a=1#

So the vertex is #(0, 3)#

#color(white)()# Alternatively, note that #x^2 >= 0# with minimum value #0# only when #x=0#. Hence the minimum value of #f(x)# is also when #x=0#. So the #x# coordinate of the vertex must be #0#.

Substituting #x=0# in the formula for #f(x)# gives us #f(0) = 3#.

Hence the vertex is at #(0, 3)#

Abigail Allen · Answer

undefined To find the vertex of the quadratic function $ f(x) = x^2 + 3 $, you use the formula $ x = \frac{-b}{2a} $ where $ a $ and $ b $ are coefficients of the quadratic equation $ f(x) = ax^2 + bx + c $. In this case, $ a = 1 $ and $ b = 0 $. Substitute these values into the formula to find $ x $, then plug $ x $ back into the original function to find the corresponding $ y $ value. So, $ x = \frac{-0}{2(1)} = 0 $. Now, substitute $ x = 0 $ into the original function to find $ y $. Thus, $ f(0) = (0)^2 + 3 = 3 $. Therefore, the vertex of the function is at $ (0, 3) $.

Nathan Axel · Answer

undefined To find the vertex of the quadratic function $ f(x) = x^2 + 3 $, you can use the formula:

$$ x = -\frac{b}{2a} $$

where $ a $ is the coefficient of the quadratic term (in this case, $ a = 1 $), and $ b $ is the coefficient of the linear term (in this case, $ b = 0 $).

Substitute the values of $ a $ and $ b $ into the formula:

$$ x = -\frac{0}{2(1)} = 0 $$

Now, plug the value of $ x $ back into the function to find the $ y $-coordinate of the vertex:

$$ f(0) = (0)^2 + 3 = 3 $$

So, the vertex of the function $ f(x) = x^2 + 3 $ is $ (0, 3) $.