Question

q5
sketch a graph of a function with the following features:

1. the y - intercept of a continuous non - linear function is - 1
2. the function is negative for - 1 < x < 3
3. the function has a minimum at (2, - 2)
4. as x → ∞, f(x) → ∞ and as x → - ∞, f(x) → ∞

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 \(y=a(x - 2)^{2}-2\):

$$
LATEXBLOCK0
$$

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 \(y = 0\):

$$
LATEXBLOCK1
$$

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 \(\lim_{x\to\pm\infty}y=\infty\).