Question

Hi guys I am a little confused I know how to work it out but for -2=0 I got -2=-8 pls help

Answer 1

`\begin{array}c\cline{1-8}\vphantom{\frac12}x&-3&-2&-1&0&1&2&3\\\cline{1-8}\vphantom{\frac12}y&5&0&-3&-4&-3&0&5\\\cline{1-8}\end{array}`

Explanation:

Given quadratic equation:

`y=x^2-4`

To complete the table of values, substitute each value of x into the equation.

Remember that when we square a negative number, the result is always positive, because a negative multiplied by a negative gives a positive result.

`\begin{aligned}x=-3 \implies y&=(-3)^2-4\\y&=9-4\\y&=5\end{aligned}`

`\begin{aligned}x=-2 \implies y&=(-2)^2-4\\y&=4-4\\y&=0\end{aligned}`

`\begin{aligned}x=-1 \implies y&=(-1)^2-4\\y&=1-4\\y&=-3\end{aligned}`

`\begin{aligned}x=0 \implies y&=(0)^2-4\\y&=0-4\\y&=-4\end{aligned}`

`\begin{aligned}x=1 \implies y&=(1)^2-4\\y&=1-4\\y&=-3\end{aligned}`

`\begin{aligned}x=2 \implies y&=(2)^2-4\\y&=4-4\\y&=0\end{aligned}`

`\begin{aligned}x=3 \implies y&=(3)^2-4\\y&=9-4\\y&=5\end{aligned}`

Therefore, the completed table of values is:

`\begin{array}\cline{1-8}\vphantom{\frac12}x&-3&-2&-1&0&1&2&3\\\cline{1-8}\vphantom{\frac12}y&5&0&-3&-4&-3&0&5\\\cline{1-8}\end{array}`

To draw the graph of y = x² - 4, plot the (x, y) points from the completed table, and connect them with a smooth curve. Ensure the upward-opening parabola is symmetric about the axis of symmetry, x = 0.