Question

How do you graph the parabola `y=2x^2 - 18x +13` using vertex, intercepts?

Answer 1

vertex `(9/2, -55/2)`

y-intercept `(0, 13)`

Explanation:

Given: `f(x) = y = 2x^2 - 18x + 13`

Find the vertex:

With the equation in the form: `Ax^2 + Bx + C = 0`

vertex `(-B/(2A), f(-B/(2A)))`

`x = 18/(2*2) = 18/4 = 9/2`

`f(9/2) = 2*(9/2)^2 - 18(9/2) + 13`

`f(9/2) = 2 * 81/4 - 81 + 13`

`f(9/2) = 81/2 -68/1 * 2/2`

`f(9/2) = 81/2 - 136/2`

`f(9/2) = -55/2`

`color(blue)("vertex "(9/2, -55/2))`

Find the y-intercept by setting x = 0:

`y = 2*0^2 - 18*0 + 13 = 13`

`color (red)(y"-intercept "(0, 13))`

Find the x-intercepts by setting y = 0:

`2x^2 - 18x + 13 = 0`

Use the quadratic formula to solve since this quadratic isn't easy to factor:

`x = (-B+-sqrt(B^2 - 4AC))/(2A)`

`x = (18+-sqrt(18^2 - 4*2*13))/(2*2)`

`x = 18/4 +- sqrt(220)/4 = 9/2 +- (sqrt(4)sqrt(55))/4`

`x = 9/2 +- sqrt(55)/2`

`color (red)(x"-intercepts "(9/2 + sqrt(55)/2, 0), (9/2 - sqrt(55)/2, 0))`

graph{2x^2 - 18x + 13 [-5, 10, -30, 15]}