Changing from quadratic formula to standard form.

Question

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 $y=af(bx+c)+d$?

Answer 1

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?

Answer 2

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 $a>0$

$x<-1$

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

$a^2+3*a^2+4*a<=0$

$4*a^2+4*a<=0$

$a*(a+1)<=0$

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 $a=-1$

:EDITED

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 $f''(1)<0$

Answer 3

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 $a = -1$.