`y=x^2-4x-32`

(a) Determine whether the parabola opens up or down. (b) Identify the axis of symmetry. (c) Identify the minimum point. (d) Find the x-intercepts.

determine whether the parabola opens up or down; explain your conclusion. identify the axis of symmetry and the vertex. identify the minimum. find the x-intercepts by factoring. show how the value of the discriminant supports your conclusions of the x-intercepts by factoring.

### 1 Answer | Add Yours

`y=x^2-4x-32`

(a) The sign of a (coefficient of x^2) determines direction of the parabola. If a is positive. the parabola opens. And if negative, it opens down.

**Since a=+1, hence the paranbola opens up.**

(b) When the parabola is either upward or downward, its axis of symmetry is the x-coordinate of the vertex(h,k), where `h=-b/(2a)` . So,

`h=-b/(2a)= -(-4)/(2*1)=2`

**Thus, the axis of symmetry is `x=2` .**

(c) The minimum point of the parabola is the vertex (h,k). To solve for k, substitute the value of h to y=x^2-4x-32.

`k=2^2-4(2)-32=4-8-32=-36`

**Hence, the minimum point is (2,-36).**

(d) To determine the x-intercept, set y=0.

`0=x^2-4x-32`

`0=(x-8)(x+4)`

Set each factor to zero and solve for x.

`x-8=0` and `x+4=0`

`x=8` `x=-4`

**So, the x-intercepts are (-4,0) and (8,0).**

Note that the discriminant determines the type of solutions the quadratic equation has if y is zero. If the discriminant is greater than zero, the solution are two real numbers. If it is less than zero, the solution is an imaginary number. And if it is equal to zero, there is only one real number as its solution.

Substituting the values of a (cofficient of x^2) , b (coefficient of x) and c (constant) to the formula of discriminant yields,

`Discriminant = b^2-4ac=(-4)^2-4*1*(-32)=16+128=144`

*Since the value of discriminant is greater than zero, this proves that there are two real solutions or two x-intercepts. *

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes