Question

HELP y = x2 – 2x -8.a.What are the x - intercept(s)?b.What is the y - intercept?c.Is there a highest or a lowest point?

Answer 1

From

`\begin{gathered} y=x^2-2x-8 \\ x^2-2x-8=0 \\ \text{The sum of -2x = -4x + 2x} \\ \text{The product of -8x}^2=-4x+2x \\ x^2-2x-8=0 \\ x^2-4x+2x-8=0 \\ x(x-4)+2(x-4)=0 \\ (x+2)(x-4)=0 \\ \text{Either } \\ x+2=0 \\ x=-2\text{ or} \\ x-4=0 \\ x=4 \end{gathered}`

So the x-intercepts are the roots of the equation, which is -2 or 4

Since x = -2 or x = 4

Now let's find the y-intercept

Since x-square is positive, there is a minimum value, that is there is a lowest point

We find the lowest point by finding the derivative of y with respect to x

`\begin{gathered} y=x^2-2x-8 \\ \frac{d\text{ y}}{d\text{ x}}=2x-2 \\ \\ \text{Now we set }\frac{d\text{ y}}{d\text{ x}}=0 \\ 2x-2=0 \\ 2x=2 \\ x=1 \end{gathered}`

So the lowest value for y is

`\begin{gathered} y=x^2-2x-8 \\ y=1^2-2(1)-8 \\ y=1-2-8 \\ y=-9 \end{gathered}`

So, the lowest value of y = -9