How do you find the vertex of #y = x^2-8x+7#?

Question

Abigail Allen · Accepted Answer

Complete the square to reformulate the quadratic in vertex form:

#y = x^2-8x+7 = (x-4)^2+(-9)#

So the vertex is at #(4, -9)# In most situations, we can finish the square like this: #y = a(x+b/(2a))^2+(c - (b^2)/(4a))# In particular, take note of the #b/(2a)# term, which yields the #-4#. Normally, you would simply notice the #a# and #b# terms and see what results from multiplying out #a(x-b/(2a))^2#, instead of using the above formula. However, in this case, let's just apply the formula... Since #a=1#, #b=-8#, and #c=7# in our instance, #y = x^2-8x+7# #= 1*(x+(-8)/(2*1))^2+(7-((-8)^2)/(4*1))# #=(x-4)^2+(7-64/4)# #=(x-4)^2 + (7-16)# #=(x-4)^2+(-9)# There is a quadratic in vertex form. In #y = a(x-h)^2+k#, the vertex is denoted by #(h,k)#. Thus, #(h, k) = (4, -9)# in our scenario How come this is the vertex? #(x-h)^2 >= 0# will only be zero when #x-h = 0#, or when #x = h#. Since #y = 0+k = k# when #x = h#, #(h, k)# results.

Genesis Allen · Answer

undefined To find the vertex of the quadratic function $ y = x^2 - 8x + 7 $, you can use the formula $ x = -b/(2a) $ to find the x-coordinate of the vertex. First, identify the values of $ a $ and $ b $ from the equation. In this case, $ a = 1 $ and $ b = -8 $. Then, substitute these values into the formula to find the x-coordinate of the vertex. $ x = -(-8)/(2*1) = 4 $. Once you have the x-coordinate, substitute it back into the original equation to find the y-coordinate. $ y = (4)^2 - 8(4) + 7 = 9 $. Therefore, the vertex of the function is $ (4, 9) $.