How do you find the vertex and intercepts for #y = (1/8)(x – 5)^2 - 3#?

3 Answers
Jul 22, 2018

#"see explanation"#

Explanation:

#"the equation of a parabola in "color(blue)"vertex form"# is.

#•color(white)(x)y=a(x-h)^2+k#

#"where "(h,k)" are the coordinates of the vertex and a"#
#"is a multiplier"#

#y=1/8(x-5)^2-3" is in this form"#

#color(magenta)"vertex "=(5,-3)#

#"to find y-intercept let x = 0"#

#y=25/8-24/8=1/8larrcolor(red)"y-intercept"#

#"to find x-intercepts let y = 0"#

#1/8(x-5)^2-3=0#

#1/8(x-5)^2=3rArr(x-5)^2=24#

#color(blue)"take the square root of both sides"#

#x-5=+-sqrt24=+-2sqrt6#

#"add 5 to both sides"#

#x=5+-2sqrt6larrcolor(red)"exact solutions"#

#x~~0.1,x~~9.9" to 1 dec. place"#

Vertex : #(5, -3)#,

x-intercepts : #=5\pm2\sqrt6# &

y-intercept: #=1/8#

Explanation:

Given equation:

#y=(1/8)(x-5)^2-3#

#1/8(x-5)^2=y+3#

#(x-5)^2=8(y+3)#

The above equation is in form of vertical parabola: #(x-x_1)^2=4a(y-y_1)# which has

Vertex #(x_1, y_1)\equiv(5, -3)#

Setting #x=0# in the above equation of parabola:

#(0-5)^2=8(y+3)#

#y=1/8#

The given parabola intersects the y-axis at #(0, 1/8)# &

y-intercept: #=1/8#

Similarly, setting #y=0# in above equation of parabola, we get

#(x-5)^2=8(0+3)#

#x-5=\pm\sqrt24#

#x=5\pm2\sqrt6#

The given parabola intersects the x-axis at two points

#(5\pm2\sqrt6, 0)#

x-intercepts : #=5\pm2\sqrt6#

Jul 22, 2018

The vertex is #(5,-3)#.
The y-intercept is #(0,1/8)# or #(0,0.125)#.
The x-intercepts are #(5+2sqrt6,0)# and #(5-2sqrt6,0)#.
The approximate x-intercepts are #(9.899,0)# and #(0.101,0)#.

Explanation:

#y=1/8(x-5)^2-3# is a quadratic equation in vertex form:

#y=a(x-h)^2+k#,

where:

#h=5# and #k=-3#

The vertex is #(h,k)#, which is #(5,-3)#.

To find the y-intercept, substitute #0# for #x# and solve for #y#.

#y=1/8(0-5)^2-3#

#y=1/8(25)-3#

#y=25/8-3#

Simplify #3# to #24/8#.

#y=25/8-24/8=1/8=0.125#

The y-intercept is #(0,1/8)# or #(0,0.125)#.

To find the x-intercepts, substitute #0# for #y# and solve.

#0=1/8(x-5)^2-3#

Multiply both sides by #8#.

#0=(x-5)^2-24#

Expand #(x-5)^2#.

#0=x^2-10x+25-24#

#0=x^2-10x+1#,

where:

#a=1#, #b=-10#, #c=1#

Use the quadratic formula.

#x=(-b+-sqrt(b^2-4ac))/(2a)#

Plug in the known values and solve.

#x=(-(-10)+-sqrt((-10)^2-4*1*1))/(2*1)#

#x=(10+-sqrt(96))/2#

Prime factorize #96#.

#x=(10+-sqrt(2xx2xx2xx2xx2xx3))/2#

#x=(10+-sqrt(2^2xx2^2xx2xx3))/2#

#x=(10+-2xx2sqrt(2xx3))/2#

#x=(10+-4sqrt6)/2#

Factor out the common #2#.

#x=5+-2sqrt6#

#x=5+2sqrt6#, #5-2sqrt6#

Approximate values for #x#:

#x~~9.899, 0.101#

The x-intercepts are #(5+2sqrt6,0)# and #(5-2sqrt6,0)#.

The approximate x-intercepts are #(9.899,0)# and #(0.101,0)#.

graph{y=1/8(x-5)^2-3 [-4.29, 15.71, -4.64, 5.36]}