Question

a parabola has x intercepts of -1 and 3 it passes through the point (1,6) write the equation if the parabola

Answer 1

$\boxed{f(x)=-\frac{3}{2}x^2+3x+\frac{9}{2}}$

Explanation:

The Standard Form of the equation of a parabola is given by the form:

$y=ax^2+bx+c$

As y is a function of x, then we can write:

$y=f(x)=ax^2+bx+c \\ \\ a, \ b, c \ Real \ Coefficients$

Also, we can write $f(x)$ in factored form as:

$f(x)=a(x-x_{1})(x-(x_{2}) \\ \\ x_{1} \ and \ x_{2} \ Are \ the \ roots$

In this case:

$x_{1}=-1 \\ \\ x_{2}=3$

So:

$f(x)=a(x-(-1))(x-3) \\ \\ f(x)=a(x+1)(x-3)$

We also know that the parabola passes through the point (1,6):

$f(1)=6=a(1+1)(1-3) \\ \\ Isolating \ a: \\ \\ 6=a(2)(-2) \\ \\ 6=a(-4) \\ \\ a=-\frac{6}{4} \\ \\ a=-\frac{3}{2}$

Finally, the equation of the parabola is:

$f(x)=-\frac{3}{2}(x+1)(x-3)$

Or applying Distributive Property:

$f(x)=-\frac{3}{2}\left(x^2-2x-3\right) \\ \\ \boxed{f(x)=-\frac{3}{2}x^2+3x+\frac{9}{2}}$

Learn more:

Piecewise-defined function: brainly.com/question/12169258

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