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Y=x^{2} +2x-8 determine the axis of symmetry, the vertex point and find the roots by factoring

2 Answers

3 votes

Answer:

We have axis of symmetry in
x=-1

The vertex point is
(-1, -9)

The roots by factoring are


x=-4


x=2

Explanation:

We have the function:


f(x)=x^2+2x-8

Factoring the quadratic equation:


y=(x+4)(x-2)

When
y=0


0=(x+4)(x-2)


x=-4


x=2

The vertex point is
(h, k)

From the equation, we have
a=1,\:b=2 \text{ and }\:c=-8


$h=-(b)/(2a)$


$h=-(2)/(2\cdot \:1)$


h=-1

We also got the axis of symmetry

In order to find
k just use the
$h=x_(vetex)$:


k=(-1)^2+2(-1)-8\\k=1-2-8\\k=-9

Once
a>0, we have the minimum at
(-1, -9)

User JamieGL
by
7.4k points
4 votes

Answer:

x = -1 axis of symmetry

(-1,-9) vertex

x = -4 x=2 are the roots

Explanation:

y = x^2 + 2x - 8

The axis of symmetry is

h = -b/2a

h = -2/ (2*1) = -2/2 =-1

x = -1

The x coordinate of the vertex is at the axis of symmetry

To find the y value substitute into the function

y = (-1)^2 +2(-1) -8

y = 1-2-8

y = -9

The vertex is at ( -1,-9)

Factor

What 2 number multiply to -8 and add to 2

4*-2 = -8

4+-2 = 2

y = ( x+4) (x-2)

Using the zero product property to find the roots

0 = ( x+4) (x-2)

x+4 =0 x-2 =0

x = -4 x=2 are the roots

User Joe Fletcher
by
7.9k points

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