Question

all you need is on the photo please don't do step-by-step dothat way vertex: and answer axis of symmetry: answer x-intercept :answermaximum or minimum:answermax/min value:answer y-intercept :answer

Answer 1

We have the function:

`f(x)=x^2-x`

We have to find the vertex.

We can do it by rearranging the equation into vertex form:

`\begin{gathered} \text{Vertex form:}\longrightarrow f(x)=a(x-h)^2+k \\ \text{Vertex:}\longrightarrow(h,k) \end{gathered}`

We can do it like this:

`\begin{gathered} f(x)=x^2-x \\ f(x)=x^2-2\cdot(1)/(2)\cdot x+((1)/(2))^2-((1)/(2))^2 \\ f(x)=(x^2-2\cdot(1)/(2)x+((1)/(2))^2)-(1)/(4) \\ f(x)=(x-(1)/(2))^2-(1)/(4) \\ \text{Vertex:}\longrightarrow((1)/(2),-(1)/(4))=(0.5,-0.25) \end{gathered}`

The axis of symmetry, as this is a parabola for axis y, is a vertical line that pass through the vertex.

Vertical lines are defined as x=constant, and in this case, the vertical line that is the axis od fymmetry is x=0.5.

The x-intercepts of f(x) are the roots. We can calculate them in this case by factorizing the equation:

`\begin{gathered} f(x)=x^2-x=x(x-1)=(x-0)(x-1) \\ \text{Roots:}\longrightarrow x_1=0,x_2=1 \end{gathered}`

The x-intercepts are x=0 and x=1.

As the value of the quadratic coefficient is a=1 and is positive we know that we have a concave up parabola.

This means that in the vertex we have a minimum value for the function.

The value for this minimum is f(0.5)=-0.25.

The y-intercept is the value of f(x) when x=0. We can find it by replacing x with 0 and calculate f(x):

`f(0)=0^2-0=0`

The y-intercept is y=0. We already know this point as it is a root of f(x).

Answer:

vertex: (0.5, -0.25)

axis of symmetry: x=0.5

x-intercept: x=0 and x=1

maximum or minimum: minimum

max/min value: y=-0.25

y-intercept: y=0 ​