Question

y = x^2 - 2x + 5, find the vertex, the axis of symmetry, and graph.

Answer 1

Given parabola y = x ^2 - 2x + 5

To find x  intercept substitute y  = 0 in above equation.

x ^2 - 2x + 5 = 0

Compare it to quadratic equation ax ^2 + bx +c = 0.

a  = 1 , b  = -2, c  = 5.

x  = [ - b ±√(b^2-4ac )]/2a

x  = [2±√ (4-20)]/2

x  = [2± √-16]/2

x = [2± 4i]/2

x  = 1±2i

The roots are imaginary.

There is no x intercepts of given parabola.

To find y  intercept substitute x  = 0 in parabola equation.

y  = (0)^2 - 2(0) + 5

y  = 0 + 0 + 5

y  = 5.

To find vertex of parabola

y  = x ^2 - 2x + 5

Compare it to standard equation of parabola y  = ax ^2 + bx + c.

a  = 1 , b  = -2 , c  = 5.

Axis of symmetry x = -b /2a.

x  = 2/2 = 1

Axis of symmetry x = 1

Substitute x  value in y  = x ^2 - 2x + 5.

y  = 1 -2 + 5

y  = 4

Vertex of parabola is ( x , y ) = ( 1 , 4 ).

y  intercepts are ( 0 , 5 ).

Graph :

Draw the coordinate plane.

Plot the above points.

Connect the plotted points neatly.

Then formed curve is indicating y  = x ^2 - 2x +5.

From the graph we can observe vertex of given parabola is ( 1 , 4).

Axis of symmetry x  = 1.