# How do you write the standard form of the equation of the parabola that has (3,4) vertex and (1,2) point?

Mar 7, 2017

Equation is $y = - {x}^{2} / 2 + 3 x - \frac{1}{2}$ or $x = - {y}^{2} / 2 + 4 y - 5$

#### Explanation:

As the vertex is $\left(3 , 4\right)$, the equation in vertex form could be either $y = a {\left(x - 3\right)}^{2} + 4$ or $x = a {\left(y - 4\right)}^{2} + 3$

If it is $y = a {\left(x - 3\right)}^{2} + 4$ and passes through $\left(1 , 2\right)$, we have

$2 = a {\left(1 - 3\right)}^{2} + 4$ or $2 = 4 a + 4$ i.e. $a = - \frac{1}{2}$

and equation of parabola is $y = - {\left(x - 3\right)}^{2} / 2 + 4$

or $y = - {x}^{2} / 2 + 3 x - \frac{9}{2} + 4$ i.e. $y = - {x}^{2} / 2 + 3 x - \frac{1}{2}$

If equation is considered to be $x = a {\left(y - 4\right)}^{2} + 3$ and passes through $\left(1 , 2\right)$, we have

$1 = a {\left(2 - 4\right)}^{2} + 3$ or $1 = 4 a + 3$ i.e. $a = - \frac{1}{2}$

and equation of parabola is $x = - {\left(y - 4\right)}^{2} / 2 + 3$

or $x = - {y}^{2} / 2 + 4 y - 8 + 3$ i.e. $x = - {y}^{2} / 2 + 4 y - 5$

and graphs of thetwo equation appear as follows:
graph{(y+x^2/2-3x+1/2)(x+y^2/2-4y+5)=0 [-8.79, 11.21, -2.56, 7.44]}