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4 years ago

7

Practice worksheet graphing quadratic functions in standard form

1 answer:

laila [671]4 years ago

4 0

Answer:

Standard form is Ax+Bx=C.

Step-by-step explanation:

The standard form is Ax+Bx=C, which is a way most equations are set up. However, you know that when you want to graph using slope, you use slope intercept form, which is y=mx+b. There are two ways you can get standard form into slope intercept form.

Way 1:

You solve the equation in terms of y, and you will always get it in slope intercept form, and then you graph from there.

Way 2:

You find the x and y intercepts by plugging in 0 for x if you're finding the y intercept and vice versa for y. you have to do this twice, then plot out your points.

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Galina-37 [17]

Answer:

2.02 i think

Step-by-step explanation:

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3 years ago

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Sort the data into this order:

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4 years ago

Is 24/8 a rational number

blsea [12.9K]

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Step-by-step explanation:

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3 years ago

Today Ebuka opened a new cereal box. He ate a bowl of the cereal, which was 8% of the cereal in

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Answer:

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Step-by-step explanation:

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3 years ago

Given the function how do I solve

BARSIC [14]

Answer: (a) -7 (b) $\bold{\dfrac{4}{25}}$ (c) $\bold{-\dfrac{5}{2}}$ (d) 4

Step-by-step explanation:

A) f(x) = 2x - 3 when x is between -5 and -2 (including -5 and -2)

B) f(x) = x² when x is between -2 and 2 (including 2)

C) $f(x)=-\dfrac{3}{2}x+\dfrac{7}{2}$ when x is between 2 and 5 (including 5)

a) Equation A includes x = -2

f(x) = 2x - 3

f(-2) = 2(-2) - 3

= -4 - 3

= -7

b) Equation B includes x = $-\dfrac{2}{5}$

f(x) = x²

$f\bigg(-\dfrac{2}{5}\bigg) = \bigg(-\dfrac{2}{5}\bigg)^2$

$=\dfrac{4}{25}$

c) Equation C includes x = 4

$f(x)=-\dfrac{3}{2}x+\dfrac{7}{2}$

$f(4)=-\dfrac{3}{2}(4)+\dfrac{7}{2}$

$=-\dfrac{12}{2}+\dfrac{7}{2}$

$=-\dfrac{5}{2}$

d) Try each equation to see if x falls within the values given.

A) f(x) = 2x - 3

-2.5 = 2x - 3

0.5 = 2x

0.25 = x NOT VALID since x should be between -5 and -2

B) f(x) = x²

-2.5 = x²

$\sqrt{-2.5}=x$ NOT VALID since x cannot be an imaginary number

C) $f(x)=-\dfrac{3}{2}x+\dfrac{7}{2}$

$-\dfrac{5}{2}=-\dfrac{3}{2}x+\dfrac{7}{2}$

$-\dfrac{12}{2}=-\dfrac{3}{2}x$

$-\dfrac{12}{2}\bigg(\dfrac{2}{3}\bigg)=x$

$4 = x$ VALID since x is between 2 and 5

5 0

4 years ago