Question

What change do you have to make to the graph of f (x) = 7x in order to graph the function g (x) = 7x+10?

Answer 1

To graph g(x) = 7x+10 from f(x) = 7x, shift the entire line upwards by 10 units to adjust for the new y-intercept of 10 while maintaining the same slope of 7.

Step-by-step explanation:

To graph the function g(x) = 7x+10 based on the graph of f(x) = 7x, you need to alter the y-intercept. Original function f(x) is a straight line with a slope of 7 and a y-intercept of 0. However, function g(x) has the same slope of 7 since the coefficient of x is the same, but the y-intercept is increased by 10. This means that you need to shift the entire line of f(x) upwards by 10 units on the y-axis.

Start by placing a point at (0,10) on the graph to represent the new y-intercept. Then, using the slope of 7, for every 1 unit you move to the right on the x-axis, move 7 units up to plot another point. Repeat this process for a few more points to get an accurate line, and then draw a straight line through your points to complete the graph of g(x).

Answer 2

`\bf ~~~~~~~~~~~~\textit{function transformations} \\\\\\ % templates f(x)= A( Bx+ C)+ D \\\\ ~~~~y= A( Bx+ C)+ D \\\\ f(x)= A√( Bx+ C)+ D \\\\ f(x)= A(\mathbb{R})^( Bx+ C)+ D \\\\ f(x)= A sin\left( B x+ C \right)+ D \\\\ --------------------`


`\bf \bullet \textit{ stretches or shrinks horizontally by } A\cdot B\\\\ \bullet \textit{ flips it upside-down if } A\textit{ is negative}\\ ~~~~~~\textit{reflection over the x-axis} \\\\ \bullet \textit{ flips it sideways if } B\textit{ is negative}`


`\bf ~~~~~~\textit{reflection over the y-axis} \\\\ \bullet \textit{ horizontal shift by }( C)/( B)\\ ~~~~~~if\ ( C)/( B)\textit{ is negative, to the right}\\\\ ~~~~~~if\ ( C)/( B)\textit{ is positive, to the left}\\\\ \bullet \textit{ vertical shift by } D\\ ~~~~~~if\ D\textit{ is negative, downwards}\\\\ ~~~~~~if\ D\textit{ is positive, upwards}\\\\ \bullet \textit{ period of }(2\pi )/( B)`

now, with that template in mind, let's see


`\bf \stackrel{parent}{f(x)=7x}\qquad \qquad \qquad \stackrel{transformed}{g(x)=7x\stackrel{D}{+10}}`

D = 10, upwards shift of 10 units.