how to use points to find polynomial ♡

a polynomial of degree $n$ can be determined using $n + 1$ points on the polynomial, because by substituting the coordinates in the polynomial you can create simultaneous equations and solve them.

also, if you know the coordinates of the vertex, you can substitute that into the vertex form of the equation.

standard form

a polynomial of degree n can be determined with n+1 points on the polynomial.
you just substitute the coordinates to form simultaneous equations and solve them

Find the equation of the parabola that passes through (-2,39), (5,46) and (1,6)

$y = a x^{2} + b x = c$

substitute values for coordinate

$39 = a (- 2)^{2} + b (- 2) + c$
$39 = 4 a - 2 b + c$

substitute values for coordinate

$46 = a (5)^{2} + b (5) + c$
$46 = 25 a + 5 b + c$

substitute values for coordinate
$6 = a (1)^{2} + b (1) + c$
$6 = a + b + c$

try to find values a b c
$39 = 4 a - 2 b + c$
$6 = a + b + c$
$33 = 3 a - 3 b$
$11 = a - b$

$46 = 25 a + 5 b + c$
$6 = a + b + c$
$40 = 24 a + 4 b$
$84 = 28 a$
$a = 3$

$11 = 3 - b$
$b = - 8$

$6 = 3 - 8 + c$
$c = 11$

therefore the parabola is
$y = 3 x^{2} - 8 x + 11$

2 degree vertex form

requires the turning point / point of inflection coordinate (h,k) and one other coordinate

Find the equation of the quadratic that has a turning point at (-2,4) and also passes through the point (7,2)

$f (x) = a (x - h)^{2} + k$

substitute h and k
$y = a (x + 2)^{2} + 4$

substitute x and y to get a
$2 = a (7 + 2)^{2} + 4$
$a = - \frac{2}{81}$

therefore the equation is
$y = - \frac{2}{81} (x + 2)^{2} + 4$

3 degree vertex form

requires the turning point / point of inflection coordinate (h,k) and one other coordinate

Find the equation of the cubic equation with a stationary point of inflection at (1,-3) and also passes through the point (-2,6)

$f (x) = a (x - h)^{3} + k$

substitute h and k
$y = a (x - 1)^{2} - 3$

substitute x and y to get a
$2 = a (- 2 - 1)^{3} - 3$
$a = \frac{1}{3}$

therefore the equation is
$y = \frac{1}{3} (x - 1)^{3} - 3$

2 degrees product form

requires the coordinates of two x-intercepts, and one other coordinate

$f (x) = a (x - x_{1}) (x - x_{2})$

Find the equation of a quadratic that has x-intercepts at (-2,0) and (4,0) and another point at (0,3)

substitute the x-intercept values
$y = a (x + 2) (x - 4)$

substitute x and y to get a
$3 = a (0 + 2) (0 - 4)$
$a = - \frac{3}{8}$

therefore the equation is
$y = - \frac{3}{8} (x + 2) (x - 4)$

3 degrees product form

requires the coordinates of three x-intercepts, and one other coordinate

$f (x) = a (x - x_{1}) (x - x_{2}) (x - x_{3})$

substitute the x-intercept values
a turning point that is a x-intercept can be included two times

Find the equation of a cubic equation that passes through (3,12) and has a x-intercept and turning point at (2,0)

$y = a (x - 0) (x - 2) (x - 2)$
$y = a x (x - 2)^{2}$

substitute x and y to get a
$12 = 3 a (3 - 2)^{2}$
$a = 4$

therefore the equation is
$y = 4 x (x - 2)^{2}$

4 degrees product form

requires the coordinates of four x-intercepts, and one other coordinate

$f (x) = a (x - x_{1}) (x - x_{2}) (x - x_{3}) (x - x_{4})$

substitute the x-intercept values
a turning point that is a x-intercept can be included two times
a point of inflection that is a x-intercept can be counted three times

Find the equation is a quartic equation that has y-intercept at (72.0) and two turning points on the x axis at x=1 and x=-6

$y = a (x - 1) (x - 1) (x + 6) (x + 6)$

substitute x and y to get a
$72 = a (0 - 1)^{2} (0 + 6)^{2}$
$a = 2$

thefore the equation is
$y = 2 (x - 1)^{2} (x + 6)^{2}$

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how to use points to find polynomial