How do you find its vertex, axis of symmetry, y-intercept and x-intercept for #f(x) = x^2 - 4x#?

Question

Julia Adams · Accepted Answer

Vertex is #(2,-4)#; axis of symmetry is #x-2=0#; #x#-intercepts are #0# and #4# and #y#-intercept is #0#.

This is the equation of parabola. Intercept form of equation of parabola is #f(x)=a(x-alpha)(x-beta)#, where #alpha# and #beta# are intercepts on #x#-axis. As we have #f(x)=x^2-4x=x(x-4)=(x-0)(x-4)#, #x#-intercepts are #0# and #4#. #y#-intercept is obtained by putting #x=0# and hence it is #0^2-4*0=0-0=0# and hence #y#-intercept is #0#. Vertex form of equation of parabola is #f(x)=a(x-h)^2+k#, where #(h,k)# is vertex and #x-h=0# is axis of symmetry Here we have #f(x)=x^2-4x=(x^2-4x+4)-4=(x-2)^2-4# Hence axis of symmetry is #x-2=0# and vertex is #(2,-4)#. graph{(x^2-4x-y)(x-2)((x-2)^2+(y+4)^2-0.03)=0 [-9.92, 10.08, -5.12, 4.88]}

Genesis Allen · Answer

undefined To find the vertex of a quadratic function in the form f(x) = ax^2 + bx + c, you use the formula: Vertex = (-b/2a, f(-b/2a)).
To find the axis of symmetry, use the formula: Axis of Symmetry = -b/2a.
The y-intercept is found by setting x = 0 and solving for y: f(0) = c.
To find the x-intercepts (or roots), set f(x) = 0 and solve for x using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a).