How do you write a quadratic function whose graph has the given characteristics: passes through (5,2), (0,2), (8,-6)?

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How do you write a quadratic function whose graph has the given characteristics: passes through (5,2), (0,2), (8,-6)?

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Answer 1 · 2016-11-23T01:28:41

$f (x) = - \frac{1}{3} x + \frac{5}{3} x + 2$ $f(x)=-1/3x+5/3x+2$

Explanation:

Let the function be $f (x)$ $f(x)$ . From the $(0, 2)$ $(0,2)$ and $(5, 2)$ $(5,2)$ , we can determine that $f (x) - 2 = a (x - 5) (x)$ $f(x)-2=a(x-5)(x)$ where a is the leading coefficient of the first term, because if you were to shift the function down 2, it would have zeros at 0 and 5. a is negative, given that the position of the third point $(8, - 6)$ $(8,-6)$ .
The a value is less than one.

$f (x) - 2 = a (x - 5) (x)$ $f(x)-2=a(x-5)(x)$
$f (x) = a (x^{2} - 5 x) + 2$ $f(x)=a(x^2-5x)+2$
$f (x) = a x^{2} - 5 a x + 2$ $f(x)=ax^2-5ax+2$

We know $(8, - 6)$ $(8,-6)$ is a point on $f (x)$ $f(x)$ , so plug in the point into the equation, and get a.
$- 6 = a {(8)}^{2} - 5 a (8) + 2$ $-6=a(8)^2-5a(8)+2$
$- 8 = 64 a - 40 a$ $-8=64a-40a$
$- 8 = 24 a$ $-8=24a$
$a = - \frac{1}{3}$ $a=-1/3$

Now, write the equation using the fact that $a = - \frac{1}{3}$ $a=-1/3$ .
$f (x) = - \frac{1}{3} x^{2} - 5 (- \frac{1}{3}) x + 2$ $f(x)=-1/3x^2-5(-1/3)x+2$
$f (x) = - \frac{1}{3} x + \frac{5}{3} x + 2$ $f(x)=-1/3x+5/3x+2$

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How do you write a quadratic function whose graph has the given characteristics: passes through (5,2), (0,2), (8,-6)?

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