# How to complete the square to find vertex and x-intercept of parabola y=2x^2-11x+2

Asked on by mavimalik

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The equation of a parabola in vertex form is `y = a(x – h)^2+ k` .

For the parabola `y = 2x^2 - 11x + 2` the vertex can be determined by expressing it in the form `y = a(x – h)^2+ k`

`y = 2x^2 - 11x + 2`

=> `2(x^2 - (11/2)x) + 2`

=> `2*(x^2 - 2*(11/4)*x + 121/16) + 2 - 121/8`

=> `2*(x - (11/4))^2 - 105/8`

The vertex of the parabola is `(11/4,- 105/8)`

The x-intercept is the solution of 2x^2 - 11x + 2 = 0

x1 = `11/4 + sqrt(121 - 16)/4`

=> `11/4 + sqrt 57/4`

x2 = `11/4 - sqrt 57/4`

The vertex of the parabola is `(11/4, -105/8)` and the x-intercepts are `11/4 + sqrt 57/4` and `11/4 - sqrt 57/4`

Top Answer

elekzy | Student, Grade 11 | (Level 1) Valedictorian

Posted on

Completing the square

2x^2-11+2

isolate the coefficients with x by putting them in brackets:

(2x^2-11x)+2

factor by 2

2(x^2-5.2x)+2

find haf and the square of the coefficient of x:

2(x^2-5.2x+[5.2/2]^2-[5.2/2]^2)+2

2(x^2-5.2x+67.6-67.6)+2

2(x^2-5.2x+67.6)+2(67.6)+2

2(x^2-5.2x+67.6)+135.2+2

2(x^2-5.2x+67.6)+137.2

vertex form:

2(x-2.6)^2+137.2

Vertex(h,k)= (2.6,137.2)

The x intercepts are found :

1. on the graph, at the points were the parabola touches the x-axis

2. when a quadratic expression is =0

i.e 2x^2-11x+2=0

first factor, then simplify

3. using the quadratic formula.

this question can be soved accurately using the quadratic formula,

where a=2; b=-11; and c=2

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