Sketch and determine the turning points of the curve y=x^2-x+1?

Question

Christopher Agresta · Accepted Answer

#"minimum turning point at "(1/2,3/4)# #"this is a quadratic function which can be sketched in "# #"the usual way"# #y=x^2-x+1# #"with "a=1,b=-1,c=1# #"since "a>0" then minimum turning point "uuu# #x_(color(red)"vertex")=-b/(2a)# #rArrx_(color(red)"vertex")=-(-1)/2=1/2# #rArry_(color(red)"vertex")=(1/2)^2-1/2+1=3/4# #rArr"minimum turning point at "(1/2,3/4)# #"the discriminant shows"# #Delta=b^2-4ac=1-4=-3# #"that the quadratic does not intersect the x-axis"# graph{x^2-x+1 [-10, 10, -5, 5]} #color(blue)"turning point using calculus"# #y=x^2-x+1# #rArrdy/dx=2x-1# #"for turning point equate "dy/dx=0# #rArr2x-1=0rArrx=1/2# #rArr"turning point at "(1/2,3/4)# #"to test if max/min use the "color(blue)"second derivative test"# #• " if "(d^2y)/(dx^2)" at "x=1/2>0" then minimum"# #• " if "(d^2y)/(dx^2)" at "x=1/2<0" then maximum"# #dy/dx=2x-1# #rArr(d^2y)/(dx^2)=2>0# #rArr"minimum turning point at "(1/2,3/4)#

Luke Alvarez · Answer

undefined To sketch the curve $ y = x^2 - x + 1 $ and determine its turning points, follow these steps:

1. **Sketching the Curve:**

- **Shape**: The given equation represents a quadratic function, which is a parabola.
   - **Direction**: Since the coefficient of $ x^2 $ is positive (1), the parabola opens upwards.
   - **Vertex**: The vertex of the parabola can be found using the formula $ x = -\frac{b}{2a} $ and substituting this value into the equation to find $ y $.

For $ y = x^2 - x + 1 $:
     $ a = 1 $ (coefficient of $ x^2 $)
     $ b = -1 $ (coefficient of $ x $)
     
     Vertex $ x $-coordinate:
     $ x = -\frac{-1}{2(1)} = \frac{1}{2} $
     
     Substituting $ x = \frac{1}{2} $ into the equation to find $ y $:
     $ y = \left(\frac{1}{2}\right)^2 - \frac{1}{2} + 1 $
     $ y = \frac{1}{4} - \frac{1}{2} + 1 $
     $ y = \frac{1}{4} - \frac{2}{4} + \frac{4}{4} $
     $ y = \frac{3}{4} $
     
     So, the vertex is $ \left(\frac{1}{2}, \frac{3}{4}\right) $.

2. **Turning Points**:

- Since the parabola opens upwards and the coefficient of $ x^2 $ is positive, the vertex represents the minimum turning point of the curve.

3. **Sketch**:

- Plot the vertex $ \left(\frac{1}{2}, \frac{3}{4}\right) $ on the coordinate plane.
   - Since the parabola is symmetric about its axis of symmetry, draw the parabola symmetrically around the vertex.
   - Draw the curve starting from one side of the vertex, passing through the vertex, and continuing symmetrically on the other side.

In summary, the curve $ y = x^2 - x + 1 $ is a parabola that opens upwards. Its turning point is a minimum at $ \left(\frac{1}{2}, \frac{3}{4}\right) $.