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Common Core: 8th Grade Math : Use Similar Triangles to Show Equal Slopes: CCSS.Math.Content.8.EE.B.6

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例子问题# 1:Use Similar Triangles To Show Equal Slopes: Ccss.Math.Content.8.Ee.B.6

A line has the equation 5x-9y=12 . What is the slope of this line?

Possible Answers:

$\frac{5}{9}$

$\frac{9}{5}$

$-\frac{4}{3}$

Correct answer:

$\frac{5}{9}$

Explanation:

You need to put the equation in y=mx+b form before you can easily find out its slope.

5x-9y=12

-9y=-5x+12

$y=\frac{5}{9}x-\frac{4}{3}$

Since $m=\frac{5}{9}$ , that must be the slope.

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例子问题# 1:Slope

The equation of a line is 12x-8y=6 . Find the slope of this line.

Possible Answers:

$-\frac{3}{4}$

$\frac{3}{4}$

$\frac{3}{2}$

$-\frac{3}{2}$

Correct answer:

$\frac{3}{2}$

Explanation:

To find the slope, you will need to put the equation in y=mx+b form. The value ofwill be the slope.

12x-8y=6

Subtractfrom either side:

8y=12x-6

Divide each side by:

$y=\frac{3}{2}x-\frac{3}{4}$

You can now easily identify the value of.

$m=\frac{3}{2}$

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例子问题# 1:Use Similar Triangles To Show Equal Slopes: Ccss.Math.Content.8.Ee.B.6

What is the-intercept of the graph of the function $f \left ( x\right ) = 2x^{2} - 7x + 5$ ?

Possible Answers:

$\left (0,0 \right )$

$\left ( 0, -3 \frac{1}{2}\right )$

$\left (0,-7 \right )$

$\left (0,5 \right )$

$\left (0,2 \right )$

Correct answer:

$\left (0,5 \right )$

Explanation:

The-intercept of the graph of a function is the point at which it intersects the-axis - that is, at which x = 0 . This point is $\left (0, f(0) \right )$ , so evaluate f(0) :

$f \left ( x\right ) = 2x^{2} - 7x + 5$

$f \left (0\right ) = 2 \cdot 0^{2} - 7 \cdot 0 + 5$

$f \left (0\right ) = 0- 0 + 5 = 5$

The-intercept is $\left (0,5 \right )$ .

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例子问题# 1:How To Find X Or Y Intercept

Give the-intercept, if there is one, of the graph of the equation

$y = \frac{1}{x^{2}+ x} + 3$

Possible Answers:

(0,3)

(0,4)

(0,2)

$\left ( 0, \frac{3}{2}\right )$

The graph has no-intercept.

Correct answer:

The graph has no-intercept.

Explanation:

The-intercept is the point at which the graph crosses the-axis; at this point, the-coordinate is 0, so substituteforin the equation:

$y = \frac{1}{x^{2}+ x} + 3$

$y = \frac{1}{0^{2}+0} + 3$

$y = \frac{1}{0} + 3$

However, since this expression has 0 in a denominator, it is of undefined value. This means that there is no value ofpaired with-coordinate 0, and, subsequently, the graph of the equation has no-intercept.

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例子问题# 1:How To Find X Or Y Intercept

Give the-intercept, if there is one, of the graph of the equation

$y = \frac{3}{\left | 3x-8\right |}$

Possible Answers:

$\left ( 0, -\frac{3}{8}\right )$

The graph has no-intercept.

$\left ( 0, \frac{3}{8}\right )$

$\left ( 0,- \frac{3}{5}\right )$

$\left ( 0, \frac{3}{5}\right )$

Correct answer:

$\left ( 0, \frac{3}{8}\right )$

Explanation:

The-intercept is the point at which the graph crosses the-axis; at this point, the-coordinate is 0, so substituteforin the equation:

$y = \frac{3}{\left | 3x-8\right |}$

$y = \frac{3}{\left | 3 \cdot 0 -8\right |}$

$y = \frac{3}{\left | 0 -8\right |}$

$y = \frac{3}{\left | -8\right |}$

$y = \frac{3}{ 8}$

The-intercept is $\left ( 0, \frac{3}{8}\right )$ .

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Example Question #4 :How To Find X Or Y Intercept

Give the-intercept, if there is one, of the graph of the equation

$y = 3(x-2)^{2} + 4 (x-7)$ .

Possible Answers:

(0, -16)

(0, -42)

The graph does not have a-intercept.

(0, 8)

(0, -64)

Correct answer:

(0, -16)

Explanation:

The-intercept is the point at which the graph crosses the-axis; at this point, the-coordinate is 0, so substituteforin the equation:

$y = 3(x-2)^{2} + 4 (x-7)$

$y = 3(0-2)^{2} + 4 (0-7)$

$y = 3( -2)^{2} + 4 ( -7)$

$y = 3 \cdot 4 + 4 ( -7)$

y = 12+ 4 ( -7)

y = 12+ (-28)

y = -16

The-intercept is the point (0, -16) .

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Example Question #5 :How To Find X Or Y Intercept

A line passes through (5, 2) and is perpendicular to the line of the equation 3x-4y = 17 . Give the-intercept of this line.

Possible Answers:

$\left ( -1\frac{3}{4}, 0\right )$

$\left (2 \frac{1}{3}, 0 \right )$

The line has no-intercept.

$\left ( 8\frac{2}{3}, 0\right )$

$\left (6 \frac{1}{2}, 0 \right )$

Correct answer:

$\left (6 \frac{1}{2}, 0 \right )$

Explanation:

First, find the slope of the second line 3x-4y = 17 by solving foras follows:

3x-4y = 17

3x-4y +4y - 17 = 17+4y - 17

4y = 3x-17

$\frac{4y }{4}=\frac{ 3x-17 }{4}$

$y = \frac{3}{4}x- \frac{17}{4}$

The equation is now in the slope-intercept form y = mx+ b ; the slope of the second line is the-coefficient $m= \frac{3}{4}$ .

The first line, being perpendicular to the second, has as its slope the opposite of the reciprocal of $\frac{3}{4}$ , which is $m=- \frac{4}{3}$ .

Therefore, we are looking for a line through (5, 2) with slope $m=- \frac{4}{3}$ . Using point-slope form

$y - y_{1} = m(x -x_{1})$

with

$x _{1 } =5, y _{1 } = 2,m=- \frac{4}{3}$ ,

the equation becomes

$y - 2 =- \frac{4}{3}(x -5)$ .

To find the-intercept, substitute 0 forand solve for:

$0 - 2 =- \frac{4}{3}(x -5)$

$- 2 =- \frac{4}{3}x+ \frac{20}{3}$

$- 2- \frac{20}{3} =- \frac{4}{3}x+ \frac{20}{3} - \frac{20}{3}$

$- \frac{26}{3} =- \frac{4}{3}x$

$- \frac{3} {4}\cdot\left (- \frac{26}{3} \right )=- \frac{3} {4}\cdot \left (- \frac{4}{3}x \right )$

$x= \frac{26}{4} = \frac{13}{2} = 6\frac{1}{2}$

The-intercept is the point $\left (6 \frac{1}{2}, 0 \right )$ .

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例子问题# 1:X And Y Intercept

A line passes through (5, 2) and is parallel to the line of the equation 3x-4y = 17 . Give the-intercept of this line.

Possible Answers:

$\left ( 8\frac{2}{3}, 0\right )$

$\left (6 \frac{1}{2}, 0 \right )$

$\left (2 \frac{1}{3}, 0 \right )$

$\left ( -1\frac{3}{4}, 0\right )$

The line has no-intercept.

Correct answer:

$\left (2 \frac{1}{3}, 0 \right )$

Explanation:

First, find the slope of the second line 3x-4y = 17 by solving foras follows:

3x-4y = 17

3x-4y +4y - 17 = 17+4y - 17

4y = 3x-17

$\frac{4y }{4}=\frac{ 3x-17 }{4}$

$y = \frac{3}{4}x- \frac{17}{4}$

The equation is now in the slope-intercept form y = mx+ b ; the slope of the second line is the-coefficient $m= \frac{3}{4}$ .

The first line, being parallel to the second, has the same slope.

Therefore, we are looking for a line through (5, 2) with slope $m= \frac{3}{4}$ . Using point-slope form

$y - y_{1} = m(x -x_{1})$

with

$x _{1 } =5, y _{1 } = 2,m= \frac{3}{4}$ ,

the equation becomes

$y - 2 = \frac{3}{4}(x -5)$ .

To find the-intercept, substitute 0 forand solve for:

$0 - 2 = \frac{3}{4}(x -5)$

$- 2 = \frac{3}{4} x - \frac{15}{4}$

$- 2 + \frac{15}{4}= \frac{3}{4} x - \frac{15}{4} + \frac{15}{4}$

$\frac{3}{4} x = \frac{7}{4}$

$\frac{4} {3}\cdot \frac{3}{4} x =\frac{4} {3}\cdot \frac{7}{4}$

$x = \frac{7}{3} = 2\frac{1}{3}$

The-intercept is the point $\left (2 \frac{1}{3}, 0 \right )$ .

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Example Question #7 :How To Find X Or Y Intercept

Give the-intercept of the line with slope $\frac{2}{5}$ that passes through point (5, 9) .

Possible Answers:

(0, -11)

$\left (0, -17 \frac{1}{2} \right )$

The line has no-intercept.

$\left (0, -27 \frac{1}{2} \right )$

(0, 7)

Correct answer:

(0, 7)

Explanation:

By the point-slope formula, this line has the equation

$y - y_{1} = m(x -x_{1})$

where

$x_{1} = 5, y_{1} = 9 ,m = \frac{2}{5}$

By substitution, the equation becomes

$y -9 = \frac{2}{5}(x -5)$

To find the-intercept, substitute 0 forand solve for:

$y -9 = \frac{2}{5}(0 -5)$

$y -9 = \frac{2}{5}( -5)$

y - 9 = -2

y = 7

The-intercept is the point (0, 7) .

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Example Question #8 :How To Find X Or Y Intercept

Give the-intercept of the line that passes through points (-3, 4) and (2, -3) .

Possible Answers:

$\left (0, - 8 \frac{1}{5} \right )$

The line has no-intercept.

$\left (0, - \frac{1}{5} \right )$

$\left (0, - \frac{1}{7} \right )$

$\left (0, - 5\frac{6}{7} \right )$

Correct answer:

$\left (0, - \frac{1}{5} \right )$

Explanation:

First, find the slope of the line, using the slope formula

$m = \frac{y_{2} -y_{1}}{x_{2}-x_{1}}$

setting $x_{1} = -3, y_{1} = 4, x_{2} = 2, y_{2} = -3$ :

$m = \frac{-3 -4}{2-(-3)} = \frac{-7}{5} =- \frac{7}{5}$

By the point-slope formula, this line has the equation

$y - y_{1} = m(x -x_{1})$

where

$x_{1} = -3, y_{1} = 4, m = - \frac{7}{5}$ ; the line becomes

$y - 4 = - \frac{7}{5}(x -(-3))$

$y - 4 = - \frac{7}{5}(x +3)$

To find the-intercept, substitute 0 forand solve for:

$y - 4 = - \frac{7}{5}(0+3)$

$y - 4 = - \frac{7}{5}(3)$

$y - 4 = - \frac{21}{5}$

$y - 4+ 4 = - \frac{21}{5} + 4$

$y = - \frac{1}{5}$

The-intercept is $\left (0, - \frac{1}{5} \right )$ .

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Common Core: 8th Grade Math : Use Similar Triangles to Show Equal Slopes: CCSS.Math.Content.8.EE.B.6

Study concepts, example questions & explanations for Common Core: 8th Grade Math

All Common Core: 8th Grade Math Resources

Example Questions

例子问题# 1:Use Similar Triangles To Show Equal Slopes: Ccss.Math.Content.8.Ee.B.6

例子问题# 1:Slope

例子问题# 1:Use Similar Triangles To Show Equal Slopes: Ccss.Math.Content.8.Ee.B.6

例子问题# 1:How To Find X Or Y Intercept

例子问题# 1:How To Find X Or Y Intercept

Example Question #4 :How To Find X Or Y Intercept

Example Question #5 :How To Find X Or Y Intercept

例子问题# 1:X And Y Intercept

Example Question #7 :How To Find X Or Y Intercept

Example Question #8 :How To Find X Or Y Intercept

All Common Core: 8th Grade Math Resources

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