How do you find the vertex and intercepts for #g(x) = x^2 - 9x + 2#?

1 Answer
Jan 15, 2018

Vertex: #(4.5, -18.25)#
x-intercepts: #((9 + sqrt(73))/2, 0)# and #((9 - sqrt(73))/2, 0)#
y-intercept: #(0, 2)#

Explanation:

#g(x) = x^2 - 9x + 2#

This equation is in standard form, or #y = ax^2 + bx + c#.

The vertex is the highest or lowest point on the graph, depending on the coefficient (value before the #x^2# if any).

To find the vertex, we first find the #x#-value of the vertex using this equation: #x = (-b)/(2a)#.

In our equation, we know that #a = 1# and #b = -9#, so let's plug them into the equation:
#x = (-(-9))/(2(1))#
#x = 9/2# or #4.5#

Now let's find the #y#-value of the vertex. To do so, we plug in what we got for #x# into the original equation:
#g(x) = (4.5)^2 - 9(4.5) + 2#
#g(x) = 20.25 - 40.5 + 2#
#g(x) = -18.25#

So our vertex of the equation is #(4.5, -18.25)#.


To find the x-intercepts...
We plug in #0# for the #y#-values in the equation and solve for #x#:
#0 = x^2 - 9x + 2#

Now, to solve this, we need to use the quadratic formula, which is this long equation :(
#x = (-b+-sqrt(b^2-4ac))/(2a)#
#x = (-(-9) +- sqrt((-9)^2 -4(1)(2)))/(2(1))#
#x = (9 +- sqrt(81 - 8))/2#
#x = (9 +- sqrt(73))/2#


To find the y-intercepts...
We plug in #0# for the #x#-values in the equation and solve for #y#:
#g(x) = 0^2 - 9(0) + 2#
#g(x) = 0 - 0 + 2#
#g(x) = 2#

So our y-intercept is at #(0, 2)#.


To show that the vertex and intercepts are correct, here is a graph of this equation:enter image source here

If you need more help on this type of question, feel free to watch this video:

If you need more help on quadratic formula, feel free to watch this video:

Hope this helps!