q5\nsketch a graph of a function with the following features:\n1) the y - intercept of a continuous non…

Explanation:

Step1: Analyze the function type

Since as (x\to\infty,f(x)\to\infty) and as (x\to-\infty,f(x)\to\infty), the function is a quadratic function (parabola) opening upwards (because for (y = ax^{2}+bx + c,a>0) has the property (\lim_{x\to\pm\infty}ax^{2}+bx + c=\infty)).

Step2: Use the vertex - form of a parabola

The vertex - form of a parabola is (y=a(x - h)^{2}+k), where ((h,k)) is the vertex. Given the minimum (vertex) at ((2,-2)), the equation is (y=a(x - 2)^{2}-2).

Step3: Find the (y) - intercept

The (y) - intercept occurs at (x = 0). We know that when (x = 0,y=-1). Substitute (x = 0) and (y=-1) into (y=a(x - 2)^{2}-2): [ \begin{align*} -1&=a(0 - 2)^{2}-2\ -1&=4a-2\ 4a&=1\ a&=\frac{1}{4} \end{align*} ] So the function is (y=\frac{1}{4}(x - 2)^{2}-2=\frac{1}{4}(x^{2}-4x + 4)-2=\frac{1}{4}x^{2}-x - 1).

Step4: Check the sign of the function

We can find the roots of the function by setting (y = 0): [ \begin{align*} \frac{1}{4}x^{2}-x - 1&=0\ x^{2}-4x - 4&=0 \end{align*} ] Using the quadratic formula (x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}) for (ax^{2}+bx + c = 0) (here (a = 1,b=-4,c=-4)), (x=\frac{4\pm\sqrt{16 + 16}}{2}=\frac{4\pm\sqrt{32}}{2}=\frac{4\pm4\sqrt{2}}{2}=2\pm2\sqrt{2}\approx2\pm2.83). The roots are (x_1=2 - 2\sqrt{2}\approx - 0.83) and (x_2=2 + 2\sqrt{2}\approx4.83). But we can also use the fact that the parabola (y=\frac{1}{4}(x - 2)^{2}-2) is a U - shaped curve. We can pick a test point in the interval ((-1,3)), say (x = 1). Then (y=\frac{1}{4}(1 - 2)^{2}-2=\frac{1}{4}-2=-\frac{7}{4}<0).

To sketch the graph:

• Plot the (y) - intercept at ((0,-1)).
• Plot the vertex at ((2,-2)).
• Since the parabola opens upwards ((a=\frac{1}{4}>0)), and we know the general shape of (y = ax^{2}+bx + c) (a U - shaped curve). We can also find a few more points: when (x=4), (y=\frac{1}{4}(4 - 2)^{2}-2=\frac{1}{4}\times4-2=-1) (point ((4,-1))).

Answer:

Sketch a parabola opening upwards with vertex at ((2,-2)), passing through ((0,-1)) and ((4,-1)), and negative (below the (x) - axis) for (x\in(-1,3)) (approximate values for intersection with (x=-1) and (x = 3) can be found by substituting (x=-1) and (x = 3) into (y=\frac{1}{4}(x - 2)^{2}-2). When (x=-1), (y=\frac{1}{4}(-1 - 2)^{2}-2=\frac{9}{4}-2=\frac{1}{4}>0) (but we know the function is negative in the open interval ((-1,3)) from the given condition, and the actual roots are (x=2\pm2\sqrt{2})). The graph has the (y) - intercept at ((0,-1)), vertex at ((2,-2)) and the end - behavior (\lim_{x\to\pm\infty}y=\infty).