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 x^{2}-(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(x^{2}-(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(x^{2}-(9/2)*x-9)=y

(-1)*(x^{2}-(9/2)*x-9)=y

-x^{2}+(9/2)*x +9 = y

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

2 years ago

Answered By Emmanuel A

$f\left(x\right)=3x^2+6x-10$ƒ(x)=3x^{2}+6x−10

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(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(x^{2}+2x+1)−12.5 using the expansion equation

(a-b)^{2}=a^{2}-2ab+b^{2}

then, $y=3x^2+6x-9.5$y=3x^{2}+6x−9.5

Hence the final solution is $y=3x^2+6x-10$y=3x^{2}+6x−10 approximately

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 x

^{2}-(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(x

^{2}-(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) = 12.5If you isolate and solve for a, a=-1.

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

a(x

^{2}-(9/2)*x-9)=y(-1)*(x

^{2}-(9/2)*x-9)=y-x

^{2}+(9/2)*x +9 = yThe final equation of the function is -x

^{2}+(9/2)*x +9 = y2 years ago

## Answered By Emmanuel A

$f\left(x\right)=3x^2+6x-10$ƒ (x)=3x

^{2}+6x−102 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 equationthen $6=a\left(-\frac{3}{2}-1\right)^2-12.5$6=a(−32 −1)

^{2}−12.5Now 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}−12.5then solving the equation

$y=3\left(x^2+2x+1\right)-12.5$y=3(x

^{2}+2x+1)−12.5 using the expansion equation(a-b)

^{2}=a^{2}-2ab+b^{2}then, $y=3x^2+6x-9.5$y=3x

^{2}+6x−9.5Hence the final solution is $y=3x^2+6x-10$y=3x

^{2}+6x−10 approximately