Where is the vertex in the parabola y = x^2 +2x - 5? i dont understand this i need x and y intercepts and please show work?

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Where is the vertex in the parabola y = x^2 +2x - 5? i dont understand this i need x and y intercepts and please show work?

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Answer 1 · 2018-05-07T03:14:30

Vertex (-1, -6)

Explanation:

y = x^2 + 2x - 5
The x-coordinate of vertex is given by the formula:
x = - b/(2a) = - 2/2 = - 1
The y- coordinate of Vertex is given by y(-1), when x = -1 -->
y(-1) = (-1)^2 + 2(-1) - 5 = - 6
Vertex (-1, -6)
graph{x^2 + 2x - 5 [-20, 20, -10, 10]}

Answer 2 · 2018-05-07T03:45:27

Vertex $(- 1, - 6)$ $(-1, -6)$
Y-intercept $(0, - 5)$ $(0,-5)$
x-intercept $(1.449, 0)$ $(1.449, 0)$
x-intercept $(- 3.449, 0)$ $(-3.449, 0)$

Explanation:

Given -

$y = x^{2} + 2 x - 5$ $y=x^2+2x-5$

Vertex is the point where the curve turns.
To find this point - first you have to calculate
for what value of $x$ $x$ the curve turns. Use the formula to find that.

$x = \frac{- b}{2 a}$ $x=(-b)/(2a)$

Where -
$b$ $b$ is the coefficient of $x$ $x$
$a$ $a$ is the coefficient of $x^{2}$ $x^2$

$x = \frac{- 2}{2 \times 1} = \frac{- 2}{2} = - 1$ $x=(-2)/(2xx1)=(-2)/2=-1$

When $x$ $x$ takes the value $- 1$ $-1$ the curve turns. At that point $x$ $x$ co ordinate is $- 1$ $-1$ , then what is $y$ $y$ coordinate. Plug in the $x = - 1$ $x=-1$ in the given equation.

$y = {(- 1)}^{2} + 2 (- 1) - 5 = 1 - 2 - 5 = - 6$ $y=(-1)^2+2(-1)-5=1-2-5=-6$

At point $(- 1, - 6)$ $(-1,-6)$ the curve turns. This point is vertex.

Vertex $(- 1, - 6)$ $(-1, -6)$
Look at the graph.

enter image source here

What is $y$ $y$ intercept?
It is the point at which the curve cuts the Y-axis. Look at the graph. At $(0, - 5)$ $(0, -5)$ the curve cuts the Y-axis.
How to find it out?
Find the What is the value of $y$ $y$ when $x$ $x$ takes the value $0$ $0$

At $x = 0; y = 0^{2} + 2 (0) - 5 = 0 + 0 - 5 = - 5$ $x=0; y=0^2+2(0)-5=0+0-5=-5$

At point $(0, - 5)$ $(0,-5)$ the curve cuts the Y-axis.
Y-intercept $(0, - 5)$ $(0,-5)$

What is X-intercept?

It is the point at which the curve cuts the x-axis. Look at this graph. The curve cuts the x axis at two points. Then, how to find it out. Find the value(s) of $x$ $x$ when $y = 0$ $y=0$

$x^{2} + 2 x - 5 = 0$ $x^2+2x-5=0$
Solve to find the value of $x$ $x$ [Squaring method is used]

$x^{2} + 2 x = 5$ $x^2+2x=5$
$x^{2} + 2 x + 1 = 5 + 1 = 6$ $x^2+2x+1=5+1=6$
${(x + 1)}^{2} = 6$ $(x+1)^2=6$
$x + 1 = \pm \sqrt{6} = \pm 2.449$ $x+1=+-sqrt6=+-2.449$
$x = 2.449 - 1 = 1.449$ $x=2.449-1=1.449$

x-intercept $(1.449, 0)$ $(1.449, 0)$

$x = - 2.449 - 1 = - 3.449$ $x=-2.449-1=-3.449$

x-intercept $(- 3.449, 0)$ $(-3.449, 0)$

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Where is the vertex in the parabola y = x^2 +2x - 5? i dont understand this i need x and y intercepts and please show work?

2 Answers

Explanation:

Explanation:

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