A quadratic function has zeros at - 2/3 and 6. The graph of the function passes through the point (1, –12.5). What is the equation of the function, in standard form?A. f(x)= x^2+9/2x+9
B. f(x)= x^2-9/2x-9
C. f(x)= 2x^2+9x+18D. f(x)= 2x^2-9x-18
6 years ago
Answered By Saad H
$f\left(x\right)=ax^2+bx+c$ƒ(x)=ax2+bx+c is the standard form for quadratic equations. 'a' cannot equal to zero, but 'b' and 'c' can be any number, including zero.
Since the zeros (the point at which the funtion crosses the x-axis) are $-\frac{2}{3}$−23 and 6 , and the graph passes through $\left(1,-12.5\right)$(1,−12.5) we can gain a lot of info.
We know that a quadratic funtion must have the form $f\left(x\right)=n\left(x+\frac{2}{3}\right)\left(x-6\right)$ƒ(x)=n(x+23)(x−6) in order to satisfy the zero conditions (n can be any number except 0)
Give that we know a point that the function passes through we can plug those in $-12.5=n\left(1+\frac{2}{3}\right)\left(1-6\right)=n\left(\frac{5}{3}\right)\left(-5\right)=-\frac{25}{3}n$−12.5=n(1+23)(1−6)=n(53)(−5)=−253n And then we solve for n:
6 years ago
Answered By Saad H
Thus the equation that satisfies all these conditions is $f\left(x\right)=\frac{3}{2}\left(x+\frac{2}{3}\right)\left(x-6\right)$ƒ(x)=32(x+23)(x−6) which is none of the above.
6 years ago
Answered By Mohsen F
The general form of a quadratic equation is Y=ax^2+bx+c. we need to find a, b, and c.
In general, the sum and product of two zeros in quadratic equation is -b/a and c/a respectively then:
by having the two zeros, 6 and -2/3
Equation 1) -b/a= 6+(-2/3) =16/3
Equation 2) c/a= 6*(-2/3)= -4
Also for the point (1,-12.5), we have
Equation 3) -12.5=a(1)^2+b(1)+c= a+b+c
There are three equations with three unknowns, a, b, and c which can be solved.
6 years ago
Answered By Saad H
$f\left(x\right)=ax^2+bx+c$ƒ (x)=ax2+bx+c is the standard form for quadratic equations. 'a' cannot equal to zero, but 'b' and 'c' can be any number, including zero.
Since the zeros (the point at which the funtion crosses the x-axis) are $-\frac{2}{3}$−23 and 6 , and the graph passes through $\left(1,-12.5\right)$(1,−12.5) we can gain a lot of info.
We know that a quadratic funtion must have the form $f\left(x\right)=n\left(x+\frac{2}{3}\right)\left(x-6\right)$ƒ (x)=n(x+23 )(x−6) in order to satisfy the zero conditions (n can be any number except 0)
Give that we know a point that the function passes through we can plug those in $-12.5=n\left(1+\frac{2}{3}\right)\left(1-6\right)=n\left(\frac{5}{3}\right)\left(-5\right)=-\frac{25}{3}n$−12.5=n(1+23 )(1−6)=n(53 )(−5)=−253 n And then we solve for n:
6 years ago
Answered By Saad H
Thus the equation that satisfies all these conditions is $f\left(x\right)=\frac{3}{2}\left(x+\frac{2}{3}\right)\left(x-6\right)$ƒ (x)=32 (x+23 )(x−6) which is none of the above.
6 years ago
Answered By Mohsen F
The general form of a quadratic equation is Y=ax^2+bx+c. we need to find a, b, and c.
In general, the sum and product of two zeros in quadratic equation is -b/a and c/a respectively then:
by having the two zeros, 6 and -2/3
Equation 1) -b/a= 6+(-2/3) =16/3
Equation 2) c/a= 6*(-2/3)= -4
Also for the point (1,-12.5), we have
Equation 3) -12.5=a(1)^2+b(1)+c= a+b+c
There are three equations with three unknowns, a, b, and c which can be solved.