# What is the equation of the parabola that has a vertex at  (2, 5)  and passes through point  (1,-1) ?

$y = - 6 {x}^{2} + 24 x - 19$ the standard form

${\left(x - 2\right)}^{2} = - \frac{1}{6} \left(y - 5\right)$ the vertex form

#### Explanation:

Assume the parabola opening downward because, the additional point is below the Vertex

Given Vertex at $\left(2 , 5\right)$ and passing through $\left(1 , - 1\right)$

Solve for $p$ first

Using Vertex form ${\left(x - h\right)}^{2} = - 4 p \left(y - k\right)$

${\left(1 - 2\right)}^{2} = - 4 p \left(- 1 - 5\right)$

${\left(- 1\right)}^{2} = - 4 p \left(- 6\right)$

$1 = 24 p$

$p = \frac{1}{24}$

Use now Vertex form ${\left(x - h\right)}^{2} = - 4 p \left(y - k\right)$ again with variables x and y only

${\left(x - 2\right)}^{2} = - 4 \left(\frac{1}{24}\right) \left(y - 5\right)$

${\left(x - 2\right)}^{2} = - \frac{1}{6} \left(y - 5\right)$

$- 6 \left({x}^{2} - 4 x + 4\right) + 5 = y$

$y = - 6 {x}^{2} + 24 x - 24 + 5$

$y = - 6 {x}^{2} + 24 x - 19$

kindly check the graph

graph{y=-6x^2+24x-19[-25,25,-12,12]}

Feb 11, 2016

The equation of paqrabola is $y = - 6 \cdot {x}^{2} + 24 \cdot x - 19$

#### Explanation:

The equation o0f the parabola is $y = a \cdot {\left(x - h\right)}^{2} + k$ Where (h,k) is the co-ordinates of vertex. So $y = a \cdot {\left(x - 2\right)}^{2} + 5$ Now the Parabola passes through point(1,-1) so $- 1 = a \cdot {\left(1 - 2\right)}^{2} + 5 \mathmr{and} - 1 = a + 5 \mathmr{and} a = - 6$
Now putting the value of a in the equation of parabola we get $y = - 6 {\left(x - 2\right)}^{2} + 5 \mathmr{and} y = - 6 \cdot {x}^{2} + 24 \cdot x - 19$
graph{-6x^2+24x-19 [-10, 10, -5, 5]} [Answer]