Math / Algebra

Questiont/3617073/25172869/5da69924b1221913c936b06ca517bfc4
Graphing Parabolas (Level 3) Score: 0/2 Penalty: none $\qquad$
Question Plot the given parabola on the axes. Plot the roots, the vertex and two other points. $y=x^{2}+2 x-8$
Set the scales and drag the axes to change the graph. MacBook Air

Studdy Solution

STEP 1

What is this asking? We need to draw the graph of a parabola, showing its roots, vertex, and two extra points on the graph. Watch out! Don't forget that the vertex is a very special point!
It's where the parabola changes direction.

STEP 2

1. Find the roots.
2. Find the vertex.
3. Find two more points.

STEP 3

To find the roots, we need to figure out where the parabola crosses the x-axis.
This happens when $y = 0$ .
So, we set $y$ to $0$ in our equation: $0 = x^2 + 2x - 8$

STEP 4

Now, we factor this quadratic equation.
We're looking for two numbers that multiply to $-8$ and add up to $2$ .
Those numbers are $4$ and $-2$ .
So, we can rewrite the equation as: $0 = (x + 4)(x - 2)$

STEP 5

This equation is true when either $x + 4 = 0$ or $x - 2 = 0$ .
This gives us our roots: $x = -4$ and $x = 2$ .
These are the points where our parabola intersects the x-axis!

STEP 6

The x-coordinate of the vertex is right in the middle of the two roots.
To find the middle, we average the roots: $(-4 + 2)/2 = -1$ .
So, the x-coordinate of the vertex is $x = -1$ .

STEP 7

Now, we plug this $x$ -value back into our original equation to find the y-coordinate of the vertex: $y = (-1)^2 + 2(-1) - 8$ $y = 1 - 2 - 8$ $y = -9$ So, the vertex is at $(-1, -9)$ !

STEP 8

We already have three points: the two roots $(-4, 0)$ and $(2, 0)$ , and the vertex $(-1, -9)$ .
Let's find two more!
We can pick any x-values we want and plug them into the equation.
Let's try $x = 0$ .

STEP 9

Plugging in $x = 0$ , we get: $y = (0)^2 + 2(0) - 8$ $y = -8$ So, another point on the parabola is $(0, -8)$ .

STEP 10

Let's pick one more x-value.
How about $x = 1$ ?

STEP 11

Plugging in $x = 1$ , we get: $y = (1)^2 + 2(1) - 8$ $y = 1 + 2 - 8$ $y = -5$ So, our last point is $(1, -5)$ .

STEP 12

We've found everything we need!
The roots are at $(-4, 0)$ and $(2, 0)$ .
The vertex is at $(-1, -9)$ .
And our two extra points are $(0, -8)$ and $(1, -5)$ .
Plot these five points on the graph, and draw a smooth curve through them to complete the parabola!

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Questiont/3617073/25172869/5da69924b1221913c936b06ca517bfc4
Graphing Parabolas (Level 3) Score: 0/2 Penalty: none $\qquad$
Question Plot the given parabola on the axes. Plot the roots, the vertex and two other points. $y=x^{2}+2 x-8$
Set the scales and drag the axes to change the graph. MacBook Air

Studdy Solution

STEP 1

What is this asking? We need to draw the graph of a parabola, showing its roots, vertex, and two extra points on the graph. Watch out! Don't forget that the vertex is a very special point!
It's where the parabola changes direction.

STEP 2

1. Find the roots.
2. Find the vertex.
3. Find two more points.

STEP 3

To find the roots, we need to figure out where the parabola crosses the x-axis.
This happens when $y = 0$ .
So, we set $y$ to $0$ in our equation: $0 = x^2 + 2x - 8$

STEP 4

Now, we factor this quadratic equation.
We're looking for two numbers that multiply to $-8$ and add up to $2$ .
Those numbers are $4$ and $-2$ .
So, we can rewrite the equation as: $0 = (x + 4)(x - 2)$

STEP 5

This equation is true when either $x + 4 = 0$ or $x - 2 = 0$ .
This gives us our roots: $x = -4$ and $x = 2$ .
These are the points where our parabola intersects the x-axis!

STEP 6

The x-coordinate of the vertex is right in the middle of the two roots.
To find the middle, we average the roots: $(-4 + 2)/2 = -1$ .
So, the x-coordinate of the vertex is $x = -1$ .

STEP 7

Now, we plug this $x$ -value back into our original equation to find the y-coordinate of the vertex: $y = (-1)^2 + 2(-1) - 8$ $y = 1 - 2 - 8$ $y = -9$ So, the vertex is at $(-1, -9)$ !

STEP 8

We already have three points: the two roots $(-4, 0)$ and $(2, 0)$ , and the vertex $(-1, -9)$ .
Let's find two more!
We can pick any x-values we want and plug them into the equation.
Let's try $x = 0$ .

STEP 9

Plugging in $x = 0$ , we get: $y = (0)^2 + 2(0) - 8$ $y = -8$ So, another point on the parabola is $(0, -8)$ .

STEP 10

Let's pick one more x-value.
How about $x = 1$ ?

STEP 11

Plugging in $x = 1$ , we get: $y = (1)^2 + 2(1) - 8$ $y = 1 + 2 - 8$ $y = -5$ So, our last point is $(1, -5)$ .

STEP 12

We've found everything we need!
The roots are at $(-4, 0)$ and $(2, 0)$ .
The vertex is at $(-1, -9)$ .
And our two extra points are $(0, -8)$ and $(1, -5)$ .
Plot these five points on the graph, and draw a smooth curve through them to complete the parabola!

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Studdy Solution

STEP 1

What is this asking? We need to draw the graph of a parabola, showing its roots, vertex, and two extra points on the graph. Watch out! Don't forget that the vertex is a *very* special point!It's where the parabola changes direction.

STEP 2

1. Find the roots.2. Find the vertex.3. Find two more points.

STEP 3

To find the roots, we need to figure out where the parabola crosses the x-axis.This happens when y=0y = 0y=0.So, we **set** yyy to 000 in our equation: 0=x2+2x−80 = x^2 + 2x - 80=x2+2x−8

STEP 4

Now, we **factor** this quadratic equation.We're looking for two numbers that multiply to −8-8−8 and add up to 222.Those numbers are 444 and −2-2−2.So, we can rewrite the equation as: 0=(x+4)(x−2)0 = (x + 4)(x - 2)0=(x+4)(x−2)

STEP 5

This equation is true when either x+4=0x + 4 = 0x+4=0 or x−2=0x - 2 = 0x−2=0.This gives us our **roots**: x=−4x = -4x=−4 and x=2x = 2x=2.These are the points where our parabola intersects the x-axis!

STEP 6

The x-coordinate of the vertex is right in the middle of the two roots.To find the middle, we **average** the roots: (−4+2)/2=−1(-4 + 2)/2 = -1(−4+2)/2=−1.So, the x-coordinate of the vertex is x=−1x = -1x=−1.

STEP 7

Now, we **plug** this xxx-value back into our original equation to find the y-coordinate of the vertex: y=(−1)2+2(−1)−8y = (-1)^2 + 2(-1) - 8y=(−1)2+2(−1)−8 y=1−2−8y = 1 - 2 - 8y=1−2−8y=−9y = -9y=−9So, the **vertex** is at (−1,−9)(-1, -9)(−1,−9)!

STEP 8

We already have three points: the two roots (−4,0)(-4, 0)(−4,0) and (2,0)(2, 0)(2,0), and the vertex (−1,−9)(-1, -9)(−1,−9).Let's find two more!We can pick any x-values we want and plug them into the equation.Let's try x=0x = 0x=0.

STEP 9

**Plugging** in x=0x = 0x=0, we get: y=(0)2+2(0)−8y = (0)^2 + 2(0) - 8y=(0)2+2(0)−8 y=−8y = -8y=−8So, another point on the parabola is (0,−8)(0, -8)(0,−8).

STEP 10

Let's pick one more x-value.How about x=1x = 1x=1?

STEP 11

**Plugging** in x=1x = 1x=1, we get: y=(1)2+2(1)−8y = (1)^2 + 2(1) - 8y=(1)2+2(1)−8 y=1+2−8y = 1 + 2 - 8y=1+2−8y=−5y = -5y=−5So, our last point is (1,−5)(1, -5)(1,−5).

STEP 12

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Studdy solves anything!

Start learning now

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What is this asking? We need to draw the graph of a parabola, showing its roots, vertex, and two extra points on the graph. Watch out! Don't forget that the vertex is a very special point!
It's where the parabola changes direction.

1. Find the roots.
2. Find the vertex.
3. Find two more points.

To find the roots, we need to figure out where the parabola crosses the x-axis.
This happens when $y = 0$ .
So, we set $y$ to $0$ in our equation: $0 = x^2 + 2x - 8$

Now, we factor this quadratic equation.
We're looking for two numbers that multiply to $-8$ and add up to $2$ .
Those numbers are $4$ and $-2$ .
So, we can rewrite the equation as: $0 = (x + 4)(x - 2)$

This equation is true when either $x + 4 = 0$ or $x - 2 = 0$ .
This gives us our roots: $x = -4$ and $x = 2$ .
These are the points where our parabola intersects the x-axis!

The x-coordinate of the vertex is right in the middle of the two roots.
To find the middle, we average the roots: $(-4 + 2)/2 = -1$ .
So, the x-coordinate of the vertex is $x = -1$ .

Now, we plug this $x$ -value back into our original equation to find the y-coordinate of the vertex: $y = (-1)^2 + 2(-1) - 8$ $y = 1 - 2 - 8$ $y = -9$ So, the vertex is at $(-1, -9)$ !

We already have three points: the two roots $(-4, 0)$ and $(2, 0)$ , and the vertex $(-1, -9)$ .
Let's find two more!
We can pick any x-values we want and plug them into the equation.
Let's try $x = 0$ .

Plugging in $x = 0$ , we get: $y = (0)^2 + 2(0) - 8$ $y = -8$ So, another point on the parabola is $(0, -8)$ .

Let's pick one more x-value.
How about $x = 1$ ?

Plugging in $x = 1$ , we get: $y = (1)^2 + 2(1) - 8$ $y = 1 + 2 - 8$ $y = -5$ So, our last point is $(1, -5)$ .