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dusya [7]
3 years ago
11

Tell the slope and y-intercept of the graph.

Mathematics
1 answer:
Mashcka [7]3 years ago
5 0

To find the slope, you can use the formula: \frac{rise}{run}. Start at where the graph intersects at the y-axis, which is -4. So, it rises up 2 units and goes left 3 units, which makes the slope \frac{2}{3}.

To find the y-intercept, you would start at where the graph intersects at the y-axis, which is -4 and this would be the y-intercept.

To find the equation, you can use the formula: y = mx + b. m is slope and b is the y-intercept. Plug the slope and y-intercept you found earlier. Thus, your final answer should be: y = \frac{2}{3}x-4

Please reach out to me if you are still confused. Hope this helps :)

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Write an equation for the translation of x^2 + y^2 = 49 by 7 units right and 4 units up
Tatiana [17]

Answer:

(x - 7)² + (y - 4)² = 49

Step-by-step explanation:

Given

Equation: x² + y² = 49

Required:

New Equation when translated 7 units right and 4 units up

Taking it one step at a time.

When the equation is translated 7 units right, this implies a negative unit along the x axis.

The equation becomes

(x - 7)² + y² = 49

When the equation is translated 4 units up, this implies a negative unit along the y axis.

(x - 7)² + (y - 4)² = 49

The expression can be further simplified but it's best left in the form of

(x - 7)² + (y - 4)² = 49

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1. y = x2 + 8x + 15<br> Find the zeros of the function by rewriting the function in intercept form
lubasha [3.4K]

The zeros of given function y=x^{2}+8 x+15 is – 5 and – 3

<u>Solution:</u>

\text { Given, equation is } y=x^{2}+8 x+15

We have to find the zeros of the function by rewriting the function in intercept form.

By using intercept form, we can put value of y as  to obtain zeros of function

We know that, intercept form of above equation is x^{2}+8 x+15=0

\text { Splitting } 8 x \text { as }(5+3) x \text { and } 15 \text { as } 5 \times 3

\begin{array}{l}{\rightarrow x^{2}+(5+3) x+5 \times 3=0} \\\\ {\rightarrow x^{2}+5 x+3 x+5 \times 3=0}\end{array}

Taking “x” as common from first two terms and “3” as common from last two terms

x (x + 5) + 3(x + 5) = 0

(x + 5)(x + 3) = 0

Equating to 0 we get,

x + 5 = 0 or x + 3 = 0

x = - 5 or – 3

Hence, the zeroes of the given function are – 5 and – 3

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