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A quadratic function has zeros at -3/2 and 6 the graph of the function passes through the point(1,-12.5). What is the equation of the function, in standard form.

2 years ago

Answered By Qi X

This quadratic function has zeros at -3/2 and 6. Thus, working backwards from these numbers to find the factors:

If x=a is a solution, then (x-a) is a factor for the quadratic equation, and vice versa. Considering (x-a) is a factor, then (x-a)=0 should have been a factor equation for the quadratic.

In order for there to be a zero at x=-3/2, the factor must be x-(-3/2), which turns into x+(3/2). (The factor equation must have been x+(3/2)=0 )

In order for there to be a zero at x=6, the factor must be x-6. (The factor equation must have been x-6=0)


Now that we have our two factors, (x-6) and (x+ (3/2)), we can start assembling the equation. 


First, we find a quadratic equation that has zeros at 6 and -3/2, which would be represented by: (x-6)(x+(3/2))=0

Expanding this using FOIL, we get x2-(9/2)x-9.


Next, we must make sure the quadratic equation not only has these roots, but also goes through the point (1, -12.5). With this extra point, we can ensure we are getting the exact quadratic formula.

We know that the general form is a(x2-(9/2)x-9)=y. We can plug the point (1,-12.5) into this equation and find that variable 'a'.

a((1)2 - (9/2)(1) -9) = 12.5

If you isolate and solve for a, a=-1.

Plugging this value back into the general form equation...



-x2+(9/2)*x +9 = y

The final equation of the function is -x2+(9/2)*x +9 = y



2 years ago

Answered By Emmanuel A

  $f\left(x\right)=3x^2+6x-10$ƒ (x)=3x2+6x10   

2 years ago

Answered By Emmanuel A

Using the general equation  $f\left(x\right)=a\left(x-p\right)^2+q$ƒ (x)=a(xp)2+q  or  $y=a\left(x-p\right)^2+q$y=a(xp)2+q ,then the points in which the graph pass through the vertex at  $\left(p,q\right)$(p,q) where p and q is equal to  $1$1  and  $-12.5$12.5  respectively and has the zeros at  $-\frac{3}{2}$32  and 6 respectively. Then  $x=-\frac{3}{2}$x=32   and  $y=6$y=6. Substituting in the equation

then  $6=a\left(-\frac{3}{2}-1\right)^2-12.5$6=a(32 1)212.5 

Now solving for a, then

 $6=a\left(\frac{25}{4}\right)-12.5$6=a(254 )12.5 

 $18.5=a\left(\frac{25}{4}\right)$18.5=a(254 ) 

therefore  $a=3$a=3 approximately

Now we have  $y=3\left(x-1\right)^2-12.5$y=3(x1)212.5 

then solving the equation

 $y=3\left(x^2+2x+1\right)-12.5$y=3(x2+2x+1)12.5 using the expansion equation


then,  $y=3x^2+6x-9.5$y=3x2+6x9.5 

Hence the final solution is  $y=3x^2+6x-10$y=3x2+6x10 approximately