Algebra Examples
Find the Axis of Symmetry y=-x^2+6x-15
Step 1
Rewrite the equation in vertex form.
Complete the square for .
Use the form , to find the values of , , and .
Consider the vertex form of a parabola.
Find the value of using the formula .
Substitute the values of and into the formula .
Cancel the common factor of and .
Move the negative one from the denominator of .
Find the value of using the formula .
Substitute the values of , and into the formula .
Substitute the values of , , and into the vertex form .
Set equal to the new right side.
Step 2
Use the vertex form, , to determine the values of , , and .
Step 3
Since the value of is negative, the parabola opens down.
Opens Down
Step 5
Find , the distance from the vertex to the focus.
Find the distance from the vertex to a focus of the parabola by using the following formula.
Substitute the value of into the formula.
Cancel the common factor of and .
Move the negative in front of the fraction.
Step 6
The focus of a parabola can be found by adding to the y-coordinate if the parabola opens up or down.
Substitute the known values of , , and into the formula and simplify.
Step 7
Find the axis of symmetry by finding the line that passes through the vertex and the focus.
#f(x)=-3x^2+3x-2#
The general formula for a quadratic equation is #ax^2+bx+c#.
#a=-3#
#b=3#
The graph of a quadratic equation is a parabola. A parabola has an axis of symmetry and a vertex. The axis of symmetry is a vertical line the divides the parabola into to equal halves. The line of symmetry is determined by the equation #x=(-b)/(2a)#. The vertex is the point where the parabola crosses its axis of symmetry, and is defined as a point #(x,y)#.
Axis of Symmetry
#x=(-b)/(2a)=(-3)/(2(-3))=-3/-6=1/2#
The axis of symmetry is the line #x=1/2#
Vertex
Determine the value for #y# by substituting #y# for #f(x)# and by substituting #1/2# for #x# in the equation,
#y=-3x^2+3x-2#
#y=-3(1/2)^2+3(1/2)-2#
#y=-3(1/4)+3/2-2#
#y=-3/4+3/2-2#
The common denominator is #8#.
#y=-3/4*2/2+3/2*4/4-2*8/8# =
#y=-6/8+12/8-16/8# =
#y=-10/8#
#y=-5/4#
The vertex is #(x,y)=(1/2,-5/4)#
X-Intercept
The x-intercepts are where the parabola crosses the x-axis.There are no x-intercepts for this equation because the vertex is below the x-axis and the parabola is facing downward.
Y-Intercept
The y-intercept is where the parabola crosses the y-axis. To find the y-intercept, make #x=0#, and solve the equation for #y#.
#y=-3(0)^2+3(0)-2# =
#y=-2#
The y-intercept is #-2#.
graph{y=-3x^2+3x-2 [-14, 14.47, -13.1, 1.14]}
Given:
#y=x^2-6x+5# is a quadratic equation in standard form:
#y=ax^2+bx+c,
where:
#a=1#, #b=-6#, and #c=5#.
Axis of symmetry: vertical line that separates the parabola into two equal halves, designated #x#.
#x=(-b)/(2a)#
#x=(-(-6))/(2*1)#
#x=6/2#
#x=3#
The axis of symmetry is #x=3#.
Vertex: maximum or minimum point of the parabola, #(x,y)#. Since #a>0#, the vertex will be the minimum point and the parabola will open upward.
Substitute #3# for #x# in the equation and solve for #y#.
#y=3^2-6(3)+5=-4#
The vertex is #(3,-4)#
X-intercepts: values of #x# when #y=0#
Substitute #0# for #y#. Solve for #x#.
#0=x^2-6x+5#
Find two numbers that when added equal #6# and when multiplies equal #5#. The numbers #-5# and #-1# meet the requirements.
#0=(x-5)(x-1)#
Set each binomial equal to zero.
#(x-5)=0#
#x=5#
#(x-1)=0#
#x=1#
The x-intercepts are #(5,0)# and #(1,0)#.
Y-intercept: value of #y# when #x=0#.
Substitute #0# for #x# and solve for #y#.
#y=0^2-6(0)+5#
#y=5#
The y-intercept is #(0,5)#.
Plot the points for the vertex, x-intercepts, and y-intercept. Sketch a parabola through the points. Do not connect the dots.
graph{y=x^2-6x+5 [-14.3, 14.17, -9.97, 4.27]}