Question #5eb19

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Question #5eb19

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Answer 1 · 2017-08-17T01:56:34

Vertex: $(- 1, 0)$ $(-1,0)$

Axis of symmetry: $x = - 1$ $x = -1$

$x$ $x$ -intercept: $(- 1, 0)$ $(-1,0)$

Explanation:

The vertex is $(- 1, 0)$ $(-1,0)$ . The equation is in vertex form, which looks like

$y = a {(x - h)}^{2} + k$ $y = a(x-h)^2 + k$

where $h$ $h$ is the $x$ $x$ -coordinate of the vertex and $k$ $k$ is the $y$ $y$ -coordinate. The equation above has no $k$ $k$ value, so it's just $0$ $0$ .

The axis of symmetry is $x = - 1$ $x = -1$ . The axis of symmetry is always the $x$ $x$ value of the vertex. You could also find it through the standard form of a parabola that looks like

$A x^{2} + B x + C = 0$ $Ax^2 + Bx + C =0$

by calculating $- \frac{B}{2 A}$ $-B/(2A)$ .

The $x$ $x$ -intercept is $(- 1, 0)$ $(-1,0)$ . You find this by setting $y = 0$ $y = 0$ and solving for $x$ $x$ .

The $x$ $x$ -intercept is the factors of the quadratic equation getting set equal to $0$ $0$ . For example, the $x$ $x$ -intercepts for the quadratic

$(x + 4) (x - 3) = 0$ $(x+4)(x-3) = 0$

are $(- 4, 0)$ $(-4,0)$ and $(3, 0)$ $(3,0)$ . You get this by doing

$x + 4 = 0$ $x + 4 = 0" "$ and $x - 3 = 0$ $" " x -3 = 0$

and solving for $x$ $x$ . For the equation above, there is only one different factor. That would only give one $x$ $x$ -intercept. The graph would bounce on the point $(- 1, 0)$ $(-1,0)$ due to the even multiplicity.

graph{(x+1)^2 [-13.89, 12.96, -10.4, 48]}

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Question #5eb19

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