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

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

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

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

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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: $$f(x)=-\pi(x+5)(x-\frac72)$$

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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 $a ∈ ℝ^+$.

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

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