# How do you write a quadratic function in standard form whose graph passes through points (1,0), (2,4), (0,2)?

Aug 29, 2017

$3 {x}^{2} - 5 x + 2.$

#### Explanation:

Suppose that, the reqd. quadr. fun. is,

$y = f \left(x\right) = a {x}^{2} + b x + c , \left(x \in R\right) \left(a , b , c \in \mathbb{R} , a \ne 0\right) \ldots \ldots \left(\star\right) .$

$f \text{ passes through "(1,0) rArr (x,y)=(1,0)" must satisfy } \left(\star\right) .$

$\therefore 0 = a \cdot {1}^{2} + b \cdot 1 + c , i . e . , a + b + c = 0. \ldots \ldots \left({\star}^{1}\right) .$

Similarly, $\left(2 , 4\right) \in f \Rightarrow 4 a + 2 b + c = 4. \ldots \ldots \ldots \left({\star}^{2}\right) ,$ and,

$\left(0 , 2\right) \in f \Rightarrow c = 2. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left({\star}^{3}\right) .$

$\left({\star}^{3}\right) \mathmr{and} \left({\star}^{1}\right) \Rightarrow a + b + 2 = 0 , \mathmr{and} , b = - a - 2 ,$ and,

sub.ing this in $\left({\star}^{2}\right) , 4 a + 2 \left(- a - 2\right) + 2 = 4 \Rightarrow a = 3 ,$ &, so,

$b = - a - 2 = - 3 - 2 = - 5.$

Thus, $a = 3 , b = - 5 , c = 2 ,$ give us the desired quadr. fun.

$f \left(x\right) = 3 {x}^{2} - 5 x + 2.$

Aug 29, 2017

$3 {x}^{2} - 5 x + 2 = 0$

#### Explanation:

Since you are given three points, three equations can be created, all in the form of $a {x}^{2} + b x + c = y$, which is used for quadratic functions.

So, the first equation using point $\left(1 , 0\right)$ is: $a {\left(1\right)}^{2} + b \left(1\right) + c = 0$
Which simplifies as $a + b + c = 0$
The second equation using $\left(2 , 4\right)$ is $4 a + 2 b + c = 4$
The third equation with $\left(0 , 2\right)$ is $c = 2$ since $x = 0$ so the "a" and "b" parts become 0 as well.

Now that you have three equations, you can solve for the three variables using substitution and elimination.

$a + b + c = 0$
$4 a + 2 b + c = 4$
$c = 2$

Multiplying the first equation by 2, you get $2 a + 2 b + 2 c = 0$
Subtracting this equation from $4 a + 2 b + c = 4$, this is obtained: $2 a - c = 4$
You can now substitute in your third equation, $c = 2$ into $2 a - c = 4$
You then get $2 a - 2 = 4$ which simplifies to $2 a = 6$ so
$a = 3$

Now that you know what "a" and "c" are, you can plug this into the second equation.
So, $12 + 2 b + 2 = 4$
$b = - 5$

One more step!
Your final equation is in the form of $a {x}^{2} + b x + c = 0$
Just put all the values into the corresponding places, and you get:
$3 {x}^{2} - 5 x + 2 = 0$