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WARRIOR [948]
2 years ago
6

Im kinda confused on this. i need help.​

Mathematics
2 answers:
RideAnS [48]2 years ago
4 0

<u>Given </u><u>:</u><u>-</u>

  • A graph is given to us .

<u>To </u><u>Find</u><u> </u><u>:</u><u>-</u>

  • The equation in slope intercept form .

<u>Answer</u><u> </u><u>:</u><u>-</u>

From the given graph , we can see that the line passes through y axis at (0,5) . So the y intercept is 5 . And the slope of the line is 5/2 = 2.5 . So ,

\sf\longrightarrow y intercept = 5 .

\sf\longrightarrow slope = 5/2 .

Now here we can use the slope intercept form as ,

\sf\longrightarrow y = mx + c

\sf\longrightarrow y = 5/2x + 5

<u>Hence</u><u> the</u><u> required</u><u> answer</u><u> is</u><u> </u><u>y </u><u>=</u><u> </u><u>5</u><u>/</u><u>2</u><u>x</u><u> </u><u>+</u><u> </u><u>5</u><u>.</u>

butalik [34]2 years ago
3 0

Answer:

y= 5/2x+5

Step-by-step explanation:

For an equation in slope intercept form we need the slope and the y-intercept. To find the y-intercept just look at which point touches the y-axis which is (0,5). To find the slope the equation is y2-y1 / x2-x1. We take two points that passes through the line like (-2,0) and (0,5). We plug the numbers in the equation which will be 5-0/ 0-(-2) = 5/2 as our slope.

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Consider the curve y=x2+3x+5. (a) Find the slope of the secant line to the curve through the points P=(4,33) and Q=(4+h,(4+h)2+3
lawyer [7]

Answer:

a) The slope of the secant line is 11 + h

b) The slope of the curve at P is 11

c) The equation of the tangent line at P is y = 11x - 11

Step-by-step explanation:

y=x²+3x+5

(a) Find the slope of the secant line to the curve through the points P=(4,33) and Q=(4+h,(4+h)²+3(4+h)+5)

(4+h)²+3(4+h)+5 = 16 + 8h + h² + 12 + 3h + 5 = 33 + 11h + h²

slope (m) = yq - yp/xq - xp → m = 33 + 11h + h² - 33/4 + h - 4

m = 11h + h²/h = h(11+h)/h = 11 + h

The slope of the secant line is 11 + h

(b) Use your answer from part (a) to find the slope of the curve at the point P.

As P (4,33) = (4+0, 33+0) at P h is 0, so the slope at P is 11 + 0 = 11

The slope of the curve at P is 11

(c) Write an equation of the tangent line to the curve at P.

y - yp = m(x - xp) → y - 33 = 11(x - 4) → y - 33 = 11x - 44 → y = 11x - 11

The equation of the tangent line at P is y = 11x - 11

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