Question

Write the equation of a parabola with x-intercept at (3,0) and (9,0) the passes through the point (10,-7)

Answer 1

Answer:

$y=-x^2+12x-27$

Step-by-step explanation:

The general equation of the parabola is $y=ax^2+bx+c$

As the parabola passes through $(3,0),(9,0)$

$0=9a+3b+c\,\,\,...(i)\\0=81a+9b+c\,\,\,...(ii)$

Subtract (i) from (ii)

$(81a+9b+c)-(9a+3b+c)=0\\72a+6b=0\\12a+b=0\\b=-12a$

Also, the parabola passes through $(10,-7)$,

$-7=100a+10b+c\,\,\,...(iii)$

Subtract (i) from (iii)

$(100a+10b+c)-(9a+3b+c)=-7-0\\91a+7b=-7$

Put $b=-12a$

$91a+7(-12a)=-7\\91a-84a=-7\\7a=-7\\a=-1$

Put $a=-1$ in $b=-12a$

$b=-12(-1)=12$

Put $a=-1,b=12$ in (i)

$0=9a+3b+c\\0=-9+36+c\\0=27+c\\c=-27$

So, equation is $y=-x^2+12x-27$

Answer 2

The equation of the parabola is y = -x² + 12x - 27

Parabola is the locus of a point such that the same distance from a fixed line, called the directrix, and a fixed point (the focus) is the same.

The equation of a parabola is in quadratic form. Hence it is given as:

y = ax² + bx + c

At point (3, 0):

0 = a(3²) + b(3) + c

9a + 3b + c = 0      (1)

At point (9, 0):

0 = a(9²) + b(9) + c

81a + 9b + c = 0      (2)

At point (10, -7):

-7 = a(10²) + b(10) + c

100a + 10b + c = -7      (3)

Solving equations 1, 2, 3 gives: a = -1, b = 12, c = -27

The equation of the parabola is hence y = -x² + 12x - 27

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