# How do you write an equation with vertex (1,-5) and y-intercept -4 Supposing that the equation of function is a quadratic, hence the function is `f(x) = ax^2 + bx + c` .

The problem provides the information that the vertex is at (1,-5), hence `f(1) = -5` .

You need to remeber that the coordinates of vertex of parabola are `x_V...

Supposing that the equation of function is a quadratic, hence the function is `f(x) = ax^2 + bx + c` .

The problem provides the information that the vertex is at (1,-5), hence `f(1) = -5` .

You need to remeber that the coordinates of vertex of parabola are `x_V =-b/(2a)`  and `y_V = (4ac - b^2)/(4a).`

`1= -b/(2a) =gt 2a = -b`

`-20a = 4ac - b^2`

`f(1) = a+ b+c =gt a+ b+c = -5 =gt c = -5 - a - b`

`c = -5 - a + 2a =gt c = a - 5`

The graph of the function intercepts y axis at y = -4 such that:

`-4 = c`

`-4 = a - 5 =gt a = 5 -4 =gt a=1`

`-b = 2 =gt b = -2`

Hence, evaluating the function that has the vertex at (1,-5) and intercepts y axis at y=-4 yields `f(x) = x^2 - 2x - 4` .

Approved by eNotes Editorial Team