How do you find the axis of symmetry, vertex and x intercepts for #y=-3x^2+6x+1#?

Question

Kennedy Adams · Accepted Answer

#"vertex": (1, 4)#, #"axis of symmetry": x = 1#
#x-"intercepts": (1-(2sqrt(3))/3, 0), (1+(2sqrt(3))/3, 0)# To find the vertex and axis of symmetry: Put the equation in standard form: #y = a(x-h)^2 + k#, where axis of symmetry :# x = h# and #"vertex" = (h, k)# First factor the #x# terms: #y = -3(x^2 -2x) + 1 # Use completing of the square: 1. Multiply the #x#-term coefficient by #1/2# to get the completed square constant: #1/2 * -2 = -1# 2. Write the completed square: #(x -1)^2 = x^2 - 2x + 1# 3. Add the #x^2 #coefficient: #-3(x -1)^2 = -3x^2 +6x -3#. 4. Notice from step 3, an extra constant was added when completing the square: #-3#. This will need to be subtracted to keep the equation the same as the original equation: #y = -3(x -1)^2 + 1 -(-3)# 5. #y = -3(x -1)^2 + 4 # #"vertex": (1, 4)#, #"axis of symmetry": x = 1# To find #x#-intercepts: Use the quadratic equation since factoring isn't simple. Put the equation in the form: #Ax^2 + Bx +C = 0# #x = (-B +- sqrt(B^2 - 4AC))/(2A) = (-6 +- sqrt(36-4(-3)(1)))/(2(-3))# #x = (-6+- sqrt(48))/(-6) = 1 +- (4sqrt(3))/-6 = 1 +-(2sqrt(3))/-3# #x-"intercepts": (1-(2sqrt(3))/3, 0), (1+(2sqrt(3))/3, 0)#

Eli Assouline · Answer

Please see the explanation.

When given a quadratic of the form: #y = ax^2+bx+c# The equation for axis of symmetry is: #x=-b/(2a)# The x coordinate of the vertex, h, is the same as the axis of symmetry: #h = -b/(2a)# The y coordinate of the vertex, k, is the function evaluated at h: #k = y(h)# The x intercepts can be found by factoring or by using the quadratic formula: #x = (-b-sqrt(b^2-4(a)(c)))/(2a)# and #x = (-b+sqrt(b^2-4(a)(c)))/(2a)# Now, to the given equation: #y = -3x^2+6x+1# Please observe that #a = -3, b = 6, and c = 1# The computation for the axis of symmetry is as follows: #x = -b/(2a)# #x = -6/(2(-3))# #x = 1 larr# the equation for the axis of symmetry. The x coordinate of the vertex, h, is the same as the axis of symmetry: #h = 1# The y coordinate of the vertex, k, is the function evaluated at h: #k = y(1)= -3(1)^2+6(1)+1# #k = 4# The vertex is the point #(1,4)# As a prelude to finding the x intercepts, compute #b^2-4(a)(c)#: #6^2-4(-3)(1) = 36+12 = 48# Because 48 is a positive real number, we know that there are two distinct roots and, because 48 is not a perfect square, we know that the quadratic will not factor. Therefore, we find x intercepts using the quadratic formula: #x = (-b-sqrt(b^2-4(a)(c)))/(2a)# and #x = (-b+sqrt(b^2-4(a)(c)))/(2a)# #x= (-6-sqrt(48))/(2(-3))# and #x= (-6+sqrt(48))/(2(-3))# #x= 1-sqrt(48/36)# and #x= 1+sqrt(48/36)# #x= 1-sqrt(4/3)# and #x= 1+sqrt(4/3)larr# x intercepts

Isaiah Anthony · Answer

undefined Axis of symmetry: $ x = \frac{-b}{2a} = \frac{-6}{2*(-3)} = 1 $

Vertex: Substitute $ x = 1 $ into the equation to find the y-coordinate: $ y = -3(1)^2 + 6(1) + 1 = 4 $
Vertex is at $ (1, 4) $

To find x-intercepts: Set $ y = 0 $ and solve for $ x $ using the quadratic formula:
$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$
$$ x = \frac{-6 \pm \sqrt{6^2 - 4*(-3)*1}}{2*(-3)} $$
$$ x = \frac{-6 \pm \sqrt{36 + 12}}{-6} $$
$$ x = \frac{-6 \pm \sqrt{48}}{-6} $$
$$ x = \frac{-6 \pm 4\sqrt{3}}{-6} $$
$$ x = 1 \pm \frac{2\sqrt{3}}{3} $$

The x-intercepts are $ x = 1 + \frac{2\sqrt{3}}{3} $ and $ x = 1 - \frac{2\sqrt{3}}{3} $.