# Basic Function Question

Can someone please help me , I’ve no idea how to do this:

Give an example of a quadratic function $$f$$ that satisfies $$f(x) ≤ 0 ⇔ x ∈ (−∞,−5) ∪ (\frac{7}{2},∞)$$.

• If the domain of $f$ is all of $\Bbb R$, this is impossible – Hagen von Eitzen May 12 at 15:14
• make $f(x)>0$ on $(-5,\frac72)$ – J. W. Tanner May 12 at 15:14
• Look at the roots you can get! $x=-5, \frac{7}{2}$, then factorise into brackets and you have your quadratic equation! – Olly Reynolds May 12 at 15:17
• Did you mean $f(x)\color{red}<0$ ? – J. W. Tanner May 12 at 15:55

Hint:

Remember a quadratic (real-valued) function has the sign of its leading coefficient, except between its roots, if any.

There no such quadratic instead $$f(x)<0$$. If it is $$f(x)<0$$. Here is a simple way. Choose a quadratic with $$-5$$ and $$\frac72$$ as roots e.g $$(x+5)(x-\frac72)$$. Now since $$f$$ is negative, multiply it by a negative constant e.g: $$f(x)=-\pi(x+5)(x-\frac72)$$

let $$f(x)$$ be $$a(x-b)(x-c)$$, where $$a, b, c ∈ ℝ$$.

Firstly the bounds indicate that the roots are $$-5$$ and $$\frac{7}{2}$$, so

$$f(x) = a(x+5)(x-\frac{7}{2})$$

Next, when $$a<0$$, the graph would be a "sad" curve, showing that the conditions are satisfied.

So it is actually any $$f(x) = a(x+5)(x-\frac{7}{2})$$ where $$a ∈ ℝ^+$$.

I'm picturing the graph of $$f(x)$$ vs. $$x$$ as a concave downward parabola

that intersects the $$x$$-axis when $$x=-5$$ and $$\frac72$$.

The peak of the parabola occurs when $$x= \frac12(-5+\frac72)=-\frac34$$.

So one example is $$-(x+\frac34)^2+c$$.

To solve for $$c$$, set $$-(-5+\frac34)^2+c=0$$ [or $$-(\frac{7}2+\frac34)^2+c=0$$] and find $$c=(\frac{17}4)^2=\frac{289}{16}.$$

So my example is $$-(x+\frac34)^2+\frac{289}{ 16}=-x^2-\frac32x+\frac{35} 2.$$