How do you find the vertex and the intercepts for #y=x^2+6x+5#?

Question

Genesis Allen · Accepted Answer

Here is what I would do:

#y# intecept When #x = 0#, we get #y = 5# Vertex and #x# Intercepts Note that the following method will not work if the equation has no #x# intercepts. Also note that there are other ways to find the vertex. #x# Intercepts x^2+6x+5 = 0# #(x+1)(x+5) = 0# #(x+1) = 0# # " "# OR #" "# #(x+5) =0# #x=-1# or #-5# the #x# intercepts are #-1# and #-5#. Vertex The vertex is midway between the #x# intercepts. So it is at the average of the #x# intercepts. The average is #1/2# of the sum, so the vertex is at #x = 1/2((1-)+(-5)) = (-6)/2 = -3# To find the #y# coordinate of the vertex, plug #-3# in for #x# in the equation #y = x^2+6x+5# at #x = -3# is #(-3)^2+6(-3)+5 = 9-18+5 = -9+5=-4# The vertex is #(-3,-4)#

Savannah Anderson · Answer

undefined To find the vertex of the quadratic function $ y = x^2 + 6x + 5 $, use the formula for the x-coordinate of the vertex, which is $ x = \frac{-b}{2a} $, where $ a = 1 $ and $ b = 6 $. Plugging in these values, we get $ x = \frac{-6}{2(1)} = -3 $. To find the y-coordinate of the vertex, substitute $ x = -3 $ into the equation to get $ y = (-3)^2 + 6(-3) + 5 = 4 $. So, the vertex is at the point (-3, 4).

To find the x-intercepts (or zeros), set $ y = 0 $ and solve the quadratic equation $ x^2 + 6x + 5 = 0 $ using the quadratic formula: $ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $, where $ a = 1 $, $ b = 6 $, and $ c = 5 $. Plugging in these values, we get $ x = \frac{-6 \pm \sqrt{6^2 - 4(1)(5)}}{2(1)} = \frac{-6 \pm \sqrt{36 - 20}}{2} = \frac{-6 \pm \sqrt{16}}{2} = \frac{-6 \pm 4}{2} $. Thus, $ x = -5 $ or $ x = -1 $.

To find the y-intercept, set $ x = 0 $ in the equation $ y = x^2 + 6x + 5 $, which gives $ y = 5 $. So, the y-intercept is at the point (0, 5).