Help please can you verify if this is correct pls
\[y=-1/4x+1/4\]
ok... so..looking at the graph the lines goes through the point (-1,1) right? and has a slope of - 1/4 right?
\[y =- \frac{ 1 }{ 4 }x + \frac{ 1 }{ 4 }\]
b would be what falls on the y axis yes?
hmmm if you're referring to y = mx + b b = y-intercept <-- where the graph touches the y-axis now, y = mx+b <--- slope-intercept form but you're asked to use point-slope form \(\bf \begin{array}{lllll} &x_1&y_1\\ % (a,b) &({\color{red}{ -1}}\quad ,&{\color{blue}{ 1}})\quad \end{array} \\\quad \\ % slope = m slope = {\color{green}{ m}}= \cfrac{rise}{run} \implies-\cfrac{1}{4} \\ \quad \\ % point-slope intercept y-{\color{blue}{ y_1}}={\color{green}{ m}}(x-{\color{red}{ x_1}})\qquad \textit{plug in the values}\\ \qquad \uparrow\\ \textit{point-slope form}\)
no asked to: Write the equation in slope-intercept form.
hmmm in the graph it says "point-slope" form :|
You are shown (by the graph) that the line passes thru (-1,1) and that the slope of the line is -1/4. You need to find the y-intercept, which is b in y=mx+b. Since you already know x, y and m here, substit. their values into y=mx+b. What is the y-coordinate of the given point? Answer: 1 What ist he x-coordinate of the given point? Your answer: So, your y=mx+b becomes 1=(-1/4)( ) +b. Again, subst. the x-coord. of the given point inside the ( ), above, and find b.
sorry it cut out
Indeed the problem requests "point slope form." The method I proposed is the fastest way to obtain an equation, which can then be converted to point-slope form.
is my point slope form above correct
Unfortunately, you're not using the point-slope form. What you have in that box is the slope-intercept form.
\[y=-\frac{ 1 }{ 4 }x+\frac{ 1 }{ 4 }\]
I'm getting mixed uo
well, the red "X" next to it, seems to be a bit omnious
ominous? ;)
ominous even =)
All right, Ivonne: YOU choose the method we use. Do you want to start with the point-slope form or with the slope-intercept form?
this is a sample one i did
just different numbers
Which form do YOU want to start with?
y=mx+b
That's the same form I started with. Know that?
yes
Great. What is the slope (given slope) of the line in question?
i would think the answer would be y=-1/4x+1/4
-1/4
m=-1/4
Let's work through the problem itself to confirm the answer, ok? right. So, in y=mx+b, let m=-1/4.
y=-1/4x+b
Great. Now, thru which given point does the line pass?
not sure
i guessed 1/4
given is (-1,1)
Only one point is given by its coordinates: (-1,1). what is x there? what is y there? Subst. these values into y=(-1.4)x+b.
1=(-1/4)-1+b
You have to enclose that '-1' within parentheses: 1=(-1/4)(-1)+b. You have 3 terms here. We want to get rid of the fraction -1/4. So, mult. all 3 terms by 4. simplify your result.
4(1)=? 4(-1/4)=? 4(b)=?
b=.75
That looks very reasonable when you compare it to the y-intercept on the graph. Let's assume that the equation in slope-int. form is y=(-1/4)x + (3/4). Let's change that to point-slope form. Know how?
\[y=-\frac{ 1 }{ 4 }x+\frac{ 3 }{ 4 }\]
Pt.-slope form looks like y-k=m(x-h), where (h,k) is the point in question. We don't know it. We have to find it. Actually, I'm wrong there. You already know the slope is -1/4 and that one point is (-1,1). Therefore, h=-1 and k=1. m=-1/4. Subst. these values into the above equation.
What you have typed is correct, but is not in pt-slope form. Thought you wanted pt-slope form.
If y=1 and x=-1 and slope m = -1/4, then y-k = m(x-h) becomes y-1 = (-1/4)(x-[-1]). That is already in pt-slope form.
You could simplify that a little and end up with y-1=(-1/4)(x+1).
that's it. Sorry, I have to go, but I will be back on OS later this evening. Glad to work with you.
thank you!
My pleasure! Bye!
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