# Sketch the graph of each of the following polynomial functions. Label all x intercepts, y intercepts, maxima, minima. f(x)= 1/2(x^2-2x+15) I did the quadratic formula first and got 1 + or - i...

Sketch the graph of each of the following polynomial functions. Label all x intercepts, y intercepts, maxima, minima.

f(x)= 1/2(x^2-2x+15)

I did the quadratic formula first and got 1 + or - i square root 14. I am unsure how to find the x and y intercepts.

f(x)= 1/2x(x-4)(x+7)

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We are asked to graph the function `f(x)=1/2(x^2-2x+15) ` :

The graph will be a parabola (f(x) is a quadratic polynomial), opening up (the leading coefficient is positive.)

To find the y-intercept we substitute x=0 to get y=15/2.

To find the x-intercept(s) (if any) we solve f(x)=0. As you noted, there are no real zeros for the function, thus there are no x-intercepts. (The graph will lie entirely above the x-axis.)

There cannot be a maximum, and the minimum will occur at the vertex. The x-coordinate of the vertex occurs at x=-b/2 (if the function is written in standard form `f(x)=ax^2+bx+c ` ). The x-coordinate of the vertex is 1, and the associated y-coordinate is f(1)=7; the vertex is at (1,7) and is the absolute minimum for the function.

** With calculus, f'(x)=x-1; in this case there is a minimum when f'(x)=0 so x=1 as above. The second derivative is 1 which is greater than zero for all x so the graph is concave up everywhere and the extrema at x=1 is a minimum.)

The graph: