How do you solve #y = x^2 -8x + 16# graphically and algebraically?

2 Answers
Feb 11, 2018

Answer:

see below

Explanation:

Graphically

the roots are where the graph crosses the #x-#axis.

that is when #y=0#

graph{x^2-8x+16 [-3.74, 14.04, -2.56, 6.33]}

As can be seen from the graph it touches the #x-#axis at one point only

#x=4#

Algebraically

we could use factorising., completing the square or the formula.

look for factorising first

#x^2-8x=16=0#

#(x-4)^2=(x-4)(x-4)=0#

#:. x-4=0=>x=4#

the repeated brackets show that we have repeated roots, and that the #x-#axis is a tangent to the graph.

Feb 11, 2018

#y = x^2 - 8x + 16#

Let's solve it algebraically first:

We solve this by finding the "zeros" or x-intercepts of the equation. To do this, we factor the equation, if possible.

We need to find 2 numbers (can be the same) that add up to #-8# and multiply to #16#.
#-4# works:
#-4 + -4 = -8#
#-4 * -4 = 16#

So our equation is:
#y = (x-4)(x-4)# or #y = (x-4)^2#

Since we are finding the x-intercept(s), let's set #y = 0#. So:
#0 = (x-4)^2#

To simplify this, let's root both sides by #2#:
#sqrt(0) = sqrt((x-4)^2)#

#0 = x-4#

#x = 4#

So the x-intercept is at #(4, 0)#.


To graph this, we find the vertex and the slope.

In this case, since there is only one x-intercept, that means that it is the vertex.

The slope depends on the coefficient, or the number in front of the highest degree term.
#x^2# means #1x^2#, so our slope is one.

Let's graph this now!

enter image source here

As you can see, the x-intercept is the same as the vertex.

Hope this helps!