How do you find a quadratic function that passes through the points $(1, 4)$ $(1, 4)$ , $(- 1, 6)$ $(-1, 6)$ and $(- 2, 16)$ $(-2, 16)$ ?

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How do you find a quadratic function that passes through the points $(1, 4)$ $(1, 4)$ , $(- 1, 6)$ $(-1, 6)$ and $(- 2, 16)$ $(-2, 16)$ ?

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Answer 1 · 2016-11-06T17:15:38

Function is $y = 3 x^{2} - x + 2$ $y=3x^2-x+2$

Explanation:

Let the quadratic function be $y = a x^{2} + b x + c$ $y=ax^2+bx+c$ . As it passes through $(1, 4)$ $(1,4)$ , $(- 1, 6)$ $(-1,6)$ and $- 2, 16$ $-2,16$ , we will have

$a {(1)}^{2} + b \cdot 1 + c = 4$ $a(1)^2+b*1+c=4$ i.e. $a + b + c = 4$ $a+b+c=4$ ................................(1)

$a {(- 1)}^{2} + b \cdot (- 1) + c = 6$ $a(-1)^2+b*(-1)+c=6$ i.e. $a - b + c = 6$ $a-b+c=6$ ................................(2)

$a {(- 2)}^{2} + b \cdot (- 2) + c = 6$ $a(-2)^2+b*(-2)+c=6$ i.e. $4 a - 2 b + c = 16$ $4a-2b+c=16$ ................................(3)

Adding (1) and (2) gives us $2 a + 2 c = 10$ $2a+2c=10$

i.e. $a + c = 5$ $a+c=5$ and $b = - 1$ $b=-1$ ................................(4)

Multiplying (1) by $2$ $2$ and adding it to (3), we get

$2 a + 2 b + 2 c + 4 a - 2 b + c = 2 \times 4 + 16$ $2a+2b+2c+4a-2b+c=2xx4+16$

or $6 a + 3 c = 24$ $6a+3c=24$ i.e. $2 a + c = 8$ $2a+c=8$ ................................(5)

Subtracting (4) from (5) $a = 3$ $a=3$ and hence $c = 2$ $c=2$

and function is $y = 3 x^{2} - x + 2$ $y=3x^2-x+2$
graph{3x^2-x+2 [-3, 3, -2, 18]}

Answer 2 · 2016-11-06T20:09:28

$f (x) = 3 x^{2} - x + 2$ $f(x) = 3x^2-x+2$

Explanation:

Given:

$(1, 4)$ $(1, 4)$ , $(- 1, 6), (- 2, 16)$ $(-1, 6), (-2, 16)$

We can immediately write down a formula for a quadratic that goes through these points by constructing terms for each distinct value of $x$ $x$ we want to match:

$f (x) = 4 \cdot \frac{(x - (- 1)) (x - (- 2))}{((1) - (- 1)) ((1) - (- 2))} +$ $f(x) = color(blue)(4)*((x-(color(blue)(-1)))(x-(color(blue)(-2))))/(((color(blue)(1))-(color(blue)(-1)))((color(blue)(1))-(color(blue)(-2)))) +$

$f (x) = 6 \cdot \frac{(x - (1)) (x - (- 2))}{((- 1) - (1)) ((- 1) - (- 2))} +$ $color(white)(f(x) =) color(blue)(6)*((x-(color(blue)(1)))(x-(color(blue)(-2))))/(((color(blue)(-1))-(color(blue)(1)))((color(blue)(-1))-(color(blue)(-2)))) +$

$f (x) = 16 \cdot \frac{(x - (1)) (x - (- 1))}{((- 2) - (1)) ((- 2) - (- 1))}$ $color(white)(f(x) =) color(blue)(16)*((x-(color(blue)(1)))(x-(color(blue)(-1))))/(((color(blue)(-2))-(color(blue)(1)))((color(blue)(-2))-(color(blue)(-1))))$

$f (x) = 4 \cdot \frac{(x + 1) (x + 2)}{(2) (3)} + 6 \cdot \frac{(x - 1) (x + 2)}{(- 2) (1)} + 16 \cdot \frac{(x - 1) (x + 1)}{(- 3) (- 1)}$ $color(white)(f(x)) = color(blue)(4) * ((x+1)(x+2))/((2)(3)) + color(blue)(6) * ((x-1)(x+2))/((-2)(1)) + color(blue)(16) * ((x-1)(x+1))/((-3)(-1))$

$f (x) = \frac{2}{3} (x^{2} + 3 x + 2) - 3 (x^{2} + x - 2) + \frac{16}{3} (x^{2} - 1)$ $color(white)(f(x)) = 2/3 (x^2+3x+2) - 3 (x^2+x-2) + 16/3(x^2-1)$

$f (x) = 3 x^{2} - x + 2$ $color(white)(f(x)) = 3x^2-x+2$

This works by adding together the desired multiples of quadratics that take the value $1$ $1$ at a given $x$ $x$ coordinate and $0$ $0$ at the others.

graph{(y-(3x^2-x+2))(8(x-1)^2+(y-4)^2-0.02)(8(x+1)^2+(y-6)^2-0.02)(8(x+2)^2+(y-16)^2-0.02)=0 [-5.35, 4.7, -3, 20]}

In general, given three points $(x_{1}, y_{1})$ $(x_1, y_1)$ , $(x_{2}, y_{2})$ $(x_2, y_2)$ and $(x_{3}, y_{3})$ $(x_3, y_3)$ with $x_{1}, x_{2}, x_{3}$ $x_1, x_2, x_3$ all distinct, a quadratic function through them is given by:

$f (x) = y_{1} \frac{(x - x_{2}) (x - x_{3})}{(x_{1} - x_{2}) (x_{1} - x_{3})} + y_{2} \frac{(x - x_{1}) (x - x_{3})}{(x_{2} - x_{1}) (x_{2} - x_{3})} + y_{3} \frac{(x - x_{1}) (x - x_{2})}{(x_{3} - x_{1}) (x_{3} - x_{2})}$ $f(x) = y_1 ((x-x_2)(x-x_3))/((x_1-x_2)(x_1-x_3)) + y_2 ((x-x_1)(x-x_3))/((x_2-x_1)(x_2-x_3)) + y_3 ((x-x_1)(x-x_2))/((x_3-x_1)(x_3-x_2))$

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How do you find a quadratic function that passes through the points $(1, 4)$ $(1, 4)$ , $(- 1, 6)$ $(-1, 6)$ and $(- 2, 16)$ $(-2, 16)$ ?

2 Answers

Explanation:

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How do you find a quadratic function that passes through the points (1, 4)(1,4)(1, 4), (-1, 6)(−1,6)(-1, 6) and (-2, 16)(−2,16)(-2, 16) ?

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How do you find a quadratic function that passes through the points $(1, 4)$ $(1, 4)$ , $(- 1, 6)$ $(-1, 6)$ and $(- 2, 16)$ $(-2, 16)$ ?