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Question
the graph of a linear function f passes through the point (-3, -6) and has a slope of \\(\frac{2}{3}\\). where would the zero of function f be located? plot the location of the zero on the coordinate plane.
Step1: Recall point - slope form
The point - slope form of a linear equation is $y - y_1=m(x - x_1)$, where $(x_1,y_1)$ is a point on the line and $m$ is the slope. Here, $x_1=-3$, $y_1 = - 6$ and $m=\frac{2}{3}$.
Substitute these values into the point - slope form: $y-(-6)=\frac{2}{3}(x - (-3))$, which simplifies to $y + 6=\frac{2}{3}(x + 3)$.
Step2: Simplify the equation
Expand the right - hand side: $y+6=\frac{2}{3}x+2$.
Subtract 6 from both sides to get the slope - intercept form ($y=mx + b$): $y=\frac{2}{3}x+2 - 6=\frac{2}{3}x-4$.
Step3: Find the zero of the function
The zero of the function is the value of $x$ when $y = 0$. Set $y = 0$ in the equation $y=\frac{2}{3}x-4$:
$0=\frac{2}{3}x-4$.
Add 4 to both sides: $\frac{2}{3}x=4$.
Multiply both sides by $\frac{3}{2}$: $x=4\times\frac{3}{2}=6$.
So the zero of the function is at the point $(6,0)$.
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The zero of the function $f$ is located at the point $(6,0)$ on the coordinate plane.