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How do you graph $y=-x+6$?

Last updated date: 29th Feb 2024
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IVSAT 2024
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Hint:Change of form of the given equation will give the x-intercept and y-intercept of the line $y=- x+6$. We change it to the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$ to find the x intercept, and y intercept of the line as $p$ and $q$ respectively. then we place the points on the axes and from there we draw the line on the graph.

Complete step by step solution:
We are taking the general equation of line to understand the slope and the intercept form of the line $y=-x+6$. The given equation is in the form of $y=mx+k$. m is the slope of the line. The slope of the line is $-1$.
We have to find the x-intercept, and y-intercept of the line $y=-x+6$.
For this we convert the given equation into the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$. From the form we get that the x intercept, and y intercept of the line will be$p$ and $q$ respectively. The points will be $\left( p,0 \right),\left( 0,q \right)$.
The given equation is $y=-x+6$. Converting into the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$, we get
& y=-x+6 \\
& \Rightarrow x+y=6 \\
& \Rightarrow \dfrac{x}{6}+\dfrac{y}{6}=1 \\
Therefore, the x intercept, and y intercept of the line $y=5x-3$ is 6 and 6 respectively. The axes intersecting points are $\left( 6,0 \right),\left( 0,6 \right)$.

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Note: A line parallel to the X-axis does not intersect the X-axis at any finite distance. Hence, we cannot get any finite x-intercept of such a line. Same goes for lines parallel to the Y-axis. In case of slope of a line the range of the slope is 0 to $\infty$
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