How do you find the vertex and intercepts for #x^2-10x-8y+33=0#?

Question

Julia Adams · Accepted Answer

#y_("intercept")=33/8#

Vertex#->(x,y)=(5,1)#

NO X-INTERCEPT

Moving #8y# to the other side of the = and change its sign

#x^2-10x+33=+8y#

To get #y# on its own divide both sides by 8

#(x^2)/8-10/8 x+33/8=y#

Write as:

#y=(x^2)/8-10/8 x+33/8#

Write as:

#y=1/8(x^2color(red)(-10)x)+33/8#

#color(green)(y_("intercept")=+33/8 larr" read directly off the equation")#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(green)(x_("vertex")=(-1/2)xx(color(red)(-10)) =+5)#
The above line is part of the process of completing the square.

#color(green)("The "x^2/8"is positive so the graph is of general shape "uu)#

Substituting #x=+5# gives:

#color(green)(y_("vertex")=1/8[color(white)(./.)5^2-10(5)color(white)(.)]+33/8 =1)#

#color(green)("Vertex"->(x,y)=(5,1))#

As the graph is of general shape #uu# and #y_("vertex")# is above the x-axis there is NO X-INTERCEPT

Hazel Anglin · Answer

undefined To find the vertex and intercepts for the quadratic equation $x^2 - 10x - 8y + 33 = 0$, first, rewrite the equation in the standard form of a quadratic equation, $y = ax^2 + bx + c$. Then, use the formulas for finding the vertex, x-intercepts (zeros), and y-intercept.