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3 years ago

Graph the function f(x) = x^2 + 4x - 5 on the coordinate plane

Mathematics

2 answers:

Andreyy893 years ago

7 0

Hello there!

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Hello there, I graphed it on my coordinate plane app, I hope that's what you needed.

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I’ll help with anything else you need!

I hope I helped

@X8lue83rryX

Naya [18.7K]3 years ago

6 0

Answer:

Step 1: a = 1, b = -4

Step 2: x = 2, f(2) = -9

Step 4: (5,0) and (-1,0)

Step 6: f(0) = -5

On the graph the points plotted are (-1,0) , (0,-5) , (2,-9) , and (5,0)

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Please don't answer wrong.

Tamiku [17]

Answer:

Equation of a circle

$(x-a)^2+(y-b)^2=r^2$

where:

(a, b) is the center
r is the radius

Given equation: $x^2+y^2=6.25$

Comparing the given equation with the general equation of a circle, the given equation is a circle with:

center = (0, 0)
radius = $\sqrt{6.25}=2.5$

To draw the circle, place the point of a compass on the origin. Make the width of the compass 2.5 units, then draw a circle about the origin.

Given equation: $x+y=1.5$

Rearrange the given equation to make y the subject: $y=-x+1.5$

Find two points on the line:

$x=-2 \implies -(-2)+1.5=3.5\implies (-2,3.5)$

$x=2 \implies -(2)+1.5\implies-0.5\implies (2,-0.5)$

Plot the found points and draw a straight line through them.

The points of intersection of the circle and the straight line are the solutions to the equation.

To solve this algebraically, substitute $y=-x+1.5$ into the equation of the circle to create a quadratic:

$\implies x^2+(-x+1.5)^2=6.25$

$\implies x^2+x^2-3x+2.25=6.25$

$\implies 2x^2-3x-4=0$

Now use the quadratic formula to solve for x:

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$

$\implies x=\dfrac{-(-3) \pm \sqrt{(-3)^2-4(2)(-4)}}{2(2)}$

$\implies x=\dfrac{3 \pm \sqrt{41}}{4}$

To find the coordinates of the points of intersection, substitute the found values of x into $y=-x+1.5$

$\implies y=-\left(\dfrac{3 + \sqrt{41}}{4}\right)+1.5=\dfrac{3-\sqrt{41}}{4}$

$\implies y=-\left(\dfrac{3 - \sqrt{41}}{4}\right)+1.5=\dfrac{3+\sqrt{41}}{4}$

Therefore, the two points of intersection are:

$\left(\dfrac{3 + \sqrt{41}}{4},\dfrac{3-\sqrt{41}}{4}\right) \textsf{ and }\left(\dfrac{3 - \sqrt{41}}{4},\dfrac{3+\sqrt{41}}{4}\right)$

Or as decimals to 2 d.p.:

(2.35, -0.85) and (-0.85, 2.35)

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jenyasd209 [6]

Answer:

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ratelena [41]

Y < (x - 4)(x + 2)

so the critical points are -2 and 4
(-2,-1) will be in the solution

( 0,-2) will not

(4,0) will not.
the answer is ((-2,-1)

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Answer:

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