Parabola Interval 124B35

1. **State the problem:** We are given the quadratic function $$f(x)=4x^2-4x+1$$ defined on the interval $$I=[-3,2]$$ and asked to analyze its graph shape and properties. 2. **Identify the type of function:** This is a quadratic function of the form $$f(x)=ax^2+bx+c$$ where $$a=4$$, $$b=-4$$, and $$c=1$$. 3. **Determine the direction of the parabola:** Since $$a=4>0$$, the parabola opens upward. 4. **Find the vertex:** The vertex $$x$$-coordinate is given by $$x=-\frac{b}{2a}=-\frac{-4}{2\times4}=\frac{4}{8}=\frac{1}{2}$$. Calculate the vertex $$y$$-coordinate: $$f\left(\frac{1}{2}\right)=4\left(\frac{1}{2}\right)^2 -4\left(\frac{1}{2}\right)+1=4\times\frac{1}{4} -2 +1=1 -2 +1=0$$ So the vertex is at $$\left(\frac{1}{2},0\right)$$. 5. **Symmetry:** The parabola is symmetric about the vertical line $$x=\frac{1}{2}$$. 6. **Evaluate the function at the interval endpoints:** - At $$x=-3$$: $$f(-3)=4(-3)^2 -4(-3) +1=4\times9 +12 +1=36 +12 +1=49$$ - At $$x=2$$: $$f(2)=4(2)^2 -4(2) +1=4\times4 -8 +1=16 -8 +1=9$$ 7. **Summary:** The parabola opens upward with vertex at $$\left(\frac{1}{2},0\right)$$, symmetric about $$x=\frac{1}{2}$$, and defined on $$[-3,2]$$ with endpoint values $$f(-3)=49$$ and $$f(2)=9$$. **Final answer:** The graph is a parabola opening upward with vertex at $$\left(\frac{1}{2},0\right)$$ symmetric about $$x=\frac{1}{2}$$ on the interval $$[-3,2]$$.