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The graph of a quadratic function has $x$-intercepts $-1$ and $3$ and a range consisting of all numbers less than or equal to $4$. Determine an expression for the function.

This is my problem. I know what the graph looks like, but I only know quadratic formula and every time I input that into WolframAlpha, it combines my terms to create a different graph. I looked up the answer to this problem to check, and the answer is $-x^2+2x+3$. The only form I know is $y=af(bx+c)+d$. So how would I get that answer from this question without using $y=af(bx+c)+d$?

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They want the maximum to be $4,$ this must occur at the $x$ value halfway between the intercepts, thus at $x = 1.$ So you have three points $(-1,0), \; (1,4), \; (3,0).$ – Will Jagy Aug 7 '12 at 21:15
you can define this function also by this way $y=(x+1)*(x-3)/c$ – dato datuashvili Aug 7 '12 at 21:23
@WillJagy but how do I get $-x^2+2x+3$ from those three points? – Austin Broussard Aug 7 '12 at 21:24
@dato what is $c$? I know it's $-1$ but where would I derive it from? – Austin Broussard Aug 7 '12 at 21:28
@EdGorcenski the only way I've ever learned was in $af(bx+c)+d$. – Austin Broussard Aug 7 '12 at 22:19
up vote 4 down vote accepted

You want a quadratic $$f(x)=a(x+1)(x-3)\,\,\,s.t.\,\,\,f(x)\leq 4\,\,\,\forall x\in \Bbb R\Longrightarrow$$ $$ax^2-2ax-3a-4\leq 0\,\,\,\forall x\in \Bbb R\Longrightarrow \Delta=4a^2+4a(3a+4)\leq 0\Longrightarrow $$ $$4a(a+1)\leq 0\Longrightarrow -1\leq a\leq 0$$

So choose any $\,a\,$ as above and $\,f(x)=a(x+1)(x-3)\,$ fulfills your conditions, say $$a=-1\Longrightarrow f(x)=-x^2+2x+3$$

Another way: Since the vertex of a parabola $\,y=ax^2+bx+c\,\,,\,a\neq 0\,$ , is the point $$\left(-\frac{b}{2a}\,,\,-\frac{\Delta}{4a}\right)\,\,,\,\Delta=b^2-4ac$$ and the wanted parabola obviously opens downwards (as we want a maximum point) and the vertex is its maximal point, we get with $\,f(x)=a(x+1)(x-3)=ax^2-2ax-3a\,$:

$$-\frac{\Delta}{4a}\leq 4\Longrightarrow -4a^2+12a\geq 16a\,\,(a<0\,\,\text{since the parabola opens downwards!})\Longrightarrow$$ $$4a(a+1)\leq0$$ and we get again the same solution as above.

Added: Since you want the parabola to be less than or equal to $\,4\,$ you have to take the left extreme value, i.e. $\,a=-1\,$...can you see it?

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very good job @DonAntonio – dato datuashvili Aug 7 '12 at 22:13
$ax^2-2ax-3a-4\leq 0\,\,\,\forall x\in \Bbb R\Longrightarrow \Delta=4a^2+4a(3a+4)\leq 0$ How did you get from the right side of the arrow to the left. – Austin Broussard Aug 7 '12 at 22:28
You mean in the first way? Just divide by $\,4\,$ both sides of the inequality and get $\,a^2+3a^2+4\leq 0\,$ ... – DonAntonio Aug 7 '12 at 22:34
But how did you get from $ax^2-2ax-3a-4\le0$ to $\Delta=4a^2+4a(3a+4)\le0$? – Austin Broussard Aug 7 '12 at 22:36
By definition, if we have the parabola $\,y=Ax^2+Bx+C\,$ , its discriminant is $\,\Delta=B^2-4AC\,$ . In our case, $\,A=a\,,\,B=-2a\,,\,C=-(3a+4)\,$ and from here the result. Of course, a parabola isn't positive (isn't negative) iff its discriminant isn't positive... – DonAntonio Aug 7 '12 at 22:39

As DonAntonio said, You want a quadratic $f(x)=a(x+1)(x−3)$ such that $f(x)≤4$ and $f(x) = 4$ for some $x$.

To avoid using calculus, use the fact that a parabola has its extreme value halfway between its zeros, so its extreme value is at $\frac{3+(-1)}{2} =1$.

If $a$ was positive, then $f(x)$ would get arbitrarily large for large enough $x$. So $a$ must be negative.

The value of $f$ at $x = 1$ is $a(2)(-2) = -4a$, and we want this to equal 4, so $a = -1$.

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your function could be $y=a*(x+1)*(x-3)$,but because range of y is less that $4$,you can determine all values of $a$,for which it happens,consider case $a>0$


in this case $y>0$ and $x>3$ means $y>0$ as well,if you solve

$a*x^2-2*a*x-3*a-4<=0$;you get




solution is $[-1; 0]$ because you have quadratic,$a$ must not equal to zero,we know that maximum is $4$,for this let us derivate function ,we have

$y'=2*a*x-2*a$ let $y'=0$ -->$x=1$

not put $x=1$ ,we get $y(1)=-4*a$ because it is equal to $4$,it means $a=-1$


if you would like to make sure that $x=1$ is maximum and not minimum,let take second derivative test,it says that at critical point $x=x_0$ where $f'(x_0)=0$.if $f''(x_0)<0$,then $x_0$ is maximum,else minimum,in our case $f''(1)=2*a$,but because $a<0$,it means that $f''(1)<0$

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It can't be $\,a>0\,$ as the wanted parabola has a maximal point...! – DonAntonio Aug 7 '12 at 21:49
yes sure,because parabola has maximum point $4$ – dato datuashvili Aug 7 '12 at 21:55

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