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.
3 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...
a(x2-(9/2)*x-9)=y
(-1)*(x2-(9/2)*x-9)=y
-x2+(9/2)*x +9 = y
The final equation of the function is -x2+(9/2)*x +9 = y
3 years ago
Answered By Emmanuel A
$f\left(x\right)=3x^2+6x-10$ƒ(x)=3x2+6x−10
3 years ago
Answered By Emmanuel A
Using the general equation $f\left(x\right)=a\left(x-p\right)^2+q$ƒ(x)=a(x−p)2+q or $y=a\left(x-p\right)^2+q$y=a(x−p)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)2−12.5
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(x−1)2−12.5
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
(a-b)2=a2-2ab+b2
then, $y=3x^2+6x-9.5$y=3x2+6x−9.5
Hence the final solution is $y=3x^2+6x-10$y=3x2+6x−10 approximately
3 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...
a(x2-(9/2)*x-9)=y
(-1)*(x2-(9/2)*x-9)=y
-x2+(9/2)*x +9 = y
The final equation of the function is -x2+(9/2)*x +9 = y
3 years ago
Answered By Emmanuel A
$f\left(x\right)=3x^2+6x-10$ƒ (x)=3x2+6x−10
3 years ago
Answered By Emmanuel A
Using the general equation $f\left(x\right)=a\left(x-p\right)^2+q$ƒ (x)=a(x−p)2+q or $y=a\left(x-p\right)^2+q$y=a(x−p)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)2−12.5
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(x−1)2−12.5
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
(a-b)2=a2-2ab+b2
then, $y=3x^2+6x-9.5$y=3x2+6x−9.5
Hence the final solution is $y=3x^2+6x-10$y=3x2+6x−10 approximately