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2 years ago

Find the slope and y-intercept from the following graph of a linear equation.

Mathematics

1 answer:

Murljashka [212]2 years ago

8 0

For this question, you need to find the slope-intercept form of the graph. To do this, start with two points and plug them into y2-y1 / x2-x1.

(-1,-1)(1,7)

7 - - 1 / 1 - - 1 = 8/2 = 4

Therfore, your slope is 4

Next, the y-intercept. This is the point where the graph crosses with the y. In this problem, the lines intercept at the point (0,3).

Therefore, your y-intercept is 3

Y-intercept = 3, Slope = 4

Slope-Intercept form is y = 4x + 3

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A curve is given by y=(x-a)√(x-b) for x≥b, where a and b are constants, cuts the x axis at A where x=b+1. Show that the gradient

ankoles [38]

<u>Answer:</u>

A curve is given by y=(x-a)√(x-b) for x≥b. The gradient of the curve at A is 1.

<u>Solution:</u>

We need to show that the gradient of the curve at A is 1

Here given that ,

$y=(x-a) \sqrt{(x-b)}$ --- equation 1

Also, according to question at point A (b+1,0)

So curve at point A will, put the value of x and y

$0=(b+1-a) \sqrt{(b+1-b)}$

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According to multiple rule of Differentiation,

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so, we get

${u}^{\prime}=1$

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$y^{\prime}=1 \times \sqrt{(x-b)}+(x-a) \times \frac{1}{2} \sqrt{(x-b)}$

By putting value of point A and putting value of eq 2 we get

$y^{\prime}=\sqrt{(b+1-b)}+(b+1-a) \times \frac{1}{2} \sqrt{(b+1-b)}$

$y^{\prime}=\frac{d y}{d x}=1$

Hence proved that the gradient of the curve at A is 1.

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