Question

Graph the function f(x) = (x + 1)(x – 5). Use the drop-down menus to complete the steps needed to graph the function. Identify the x-intercepts: (–1, 0) and (5, 0) Find the midpoint between the intercepts: (2, 0) Find the vertex: Find the y-intercept:

Answer 1

Vertex is (2, -9)
Y Intercept is (0, -5)

Answer 2

Answer:

1. Midpoint of the intercepts is $(2,0)$
2. Vertex is $(2, -9)$
3. y intercept is $(0,-5)$

Step-by-step explanation:

Here the function is,

$f(x) = (x + 1)(x-5)$

The x intercept is the point where the graph intersects the x axis or the point where y or f(x) is 0.

So,

$\Rightarrow f(x) = 0$

$\Rightarrow (x + 1)(x-5) = 0$

$\Rightarrow x+1=0, x-5=0$

$\Rightarrow x=-1, x=5$

Hence the x-intercepts are $(-1,0),(5,0)$

Their mid point will be,

$=(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2})$

$=(\dfrac{-1+5}{2},\dfrac{0+0}{2})$

$=(2,0)$

The general vertex form of quadratic function or parabola is,

$y=a(x-h)^2+k$

where (h, k) is the vertex.

Here,

$y= (x + 1)(x-5)$

$=x^2-5x+x-5$

$=x^2-4x-5$

$=x^2-2\cdot x\cdot 2-5$

$=x^2-2\cdot x\cdot 2+2^2-5-2^2$

$=(x-2)^2-5-4$

$=(x-2)^2-9$

$\Rightarrow y=(x-2)^2-9$

Comparing this to the general form, the vertex is $(2, -9)$

The y intercept is the point where the graph intersects the y axis or the point where x is 0.

Putting x=0 in $f(x) = (x + 1)(x-5)$, we get

$f(0) = (0 + 1)(0-5)=(1)(-5)=-5$

Therefore, the y intercept is $(0,-5)$