Linear and Quadratic Graphs

Match the following

• Factored Form (Intercept Form)

• Vertex Form

• Standard Form Quadratic

• Standard Form Linear

• Slope-Intercept Form

Tags

CCSS.MATH.CONTENT.HSF.IF.C.7.a

Answer explanation

Change the sign of the -4 to 4, keep the sign of -3 as -3. Plot the point (4, -3) and use a = 1 to go right 1 and up 1.

Tags

CCSS.MATH.CONTENT.HSF.IF.C.7.a

Answer explanation

Change the sign of the 3 to -3, keep the sign of 5 as 5. Plot the point (-3, 5) and use a = -2 to go right 1 and down 2.

Tags

CCSS.MATH.CONTENT.HSF.IF.C.7.a

Answer explanation

Plot 4 on the y-axis. Then use the slope of -3/2 to go down (because of the negative) 3 and right 2.

Tags

CCSS.MATH.CONTENT.HSF.IF.C.7.a

Answer explanation

Cover up 3y. This leaves us with -2x = 6, which solves into x = -3. Plot a dot on the x-axis at -3. Cover up -2x. This leaves us with 3y = 6, which solves into y = 2. Plot a do on the y-axis at 2.

Tags

CCSS.MATH.CONTENT.HSF.IF.C.7.a

Tags

CCSS.MATH.CONTENT.HSF.IF.C.7.a

Tags

CCSS.MATH.CONTENT.HSF.IF.C.7.a

Tags

CCSS.MATH.CONTENT.HSF.IF.C.7.a

Plot the three points you would need to graph f(x) = 2(x+2)(x+6)

Answer explanation

Remember, things inside the parenthesis changes signs. So we have a dot on the x axis at -2 and -6. We find the middle of -2 and -6, which is -4 and we plug it into the equation, which gets us -8. So we need a third point at (-4,-8)

Tags

CCSS.MATH.CONTENT.HSF.IF.C.7.a

Plot the three points you would need to graph f(x) = (x-1)(x+5)

Answer explanation

Remember, things inside the parenthesis changes signs. So we have a dot on the x axis at 1 and -5. We find the middle of 1 and -5, which is -2 and we plug it into the equation, which gets us -9. So we need a third point at (-2,-9)

Tags

CCSS.MATH.CONTENT.HSF.IF.C.7.a