Question

How do you write a quadratic function in standard form whose graph passes through points (1/2, -3/10), (1, 6/5), (1/4, -3/10)?

Answer 1

Start with the vertex form and use symmetry to find the x coordinate.
Use the points to find the other 2 two values of the vertex form.
Expand the vertex form into the standard form.

Explanation:

Because you specified a function we discard the vertex form `x = a(y-k)^2 + h` and use only the form:

`y = a(x-h)^2+k" [1]"`

Because the we know the 2 x values corresponding to y value, `-3/10` we know know that the x coordinate of the vertex is halfway between `1/4 and 1/2`

`h = (1/2+1/4)/2`

`h = 3/8`

Substitute `3/8` for h into equation [1]:

`y = a(x-3/8)^2+k" [2]"`

Substitute the point `(1/2,-3/10)` into equation [2]:

`-3/10 = a(1/2-3/8)^2+k`

`-3/10 = a(1/8)^2+k`

`-3/10 = a/64+k" [3]"`

Substitute the point `(1,6/5)` into equation [2]:

`6/5 = a(1-3/8)^2+k`

`6/5 = a(5/8)^2+k`

`6/5 = (25a)/64+k" [4]"`

Subtract equation [3] from equation [4]:

`6/5+3/10 = (24a)/64`

`15/10 = 24/64a`

`a = 3/2(64/24)`

`a = 1/2(64/8)`

`a = 4`

Use equation [3] to find the value of k:

`-3/10 = 4/64+k`

`-3/10 = 1/16+k`

`k = -3/10 - 1/16`

`k = -29/80`

Substitute the value of "a" and "k" into equation [2]:

`y = 4(x-3/8)^2-29/80`

Expand the square:

`y = 4(x^2 - 6/8x +9/64)-29/80`

Distribute the 4:

`y = 4x^2 - 3x +9/16-29/80`

This is the standard form:

`y = 4x^2 - 3x +1/5`

The following is a graph of the equation and the 3 points: