Question #576da

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Question #576da

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Answer 1 · 2017-08-25T08:55:06

$x = - 6, x = - 4, x = - 5, (- 5, - 1)$ $x=-6,x=-4,x=-5,(-5,-1)$

Explanation:

$to find the x-intercepts solve x^{2} + 10 x + 24 = 0$ $"to find the x-intercepts solve "x^2+10x+24=0$

$the factors of 24 which sum to + 10 are + 4 and + 6$ $"the factors of 24 which sum to + 10 are + 4 and + 6"$

$\Rightarrow (x + 6) (x + 4) = 0$ $rArr(x+6)(x+4)=0$

$equate each factor to zero and solve for x$ $"equate each factor to zero and solve for x"$

$x + 6 = 0 \Rightarrow x = - 6 \leftarrow x-intercept$ $x+6=0rArrx=-6larrcolor(red)" x-intercept"$

$3 x + 4 = 0 \Rightarrow x = - 4 \leftarrow x-intercept$ $3x+4=0rArrx=-4larrcolor(red)" x-intercept"$

$given the parabola in standard form a x^{2} + b x + c$ $"given the parabola in standard form "ax^2+bx+c$

$the x-coordinate of the vertex/ axis of symmetry is$ $"the x-coordinate of the vertex/ axis of symmetry is"$

$x_{vertex} = - \frac{b}{2 a}$ $x_(color(red)"vertex")=-b/(2a)$

$y = x^{2} + 10 x + 24 is in standard form$ $y=x^2+10x+24" is in standard form"$

$with a = 1, b = 10, c = 24$ $"with "a=1,b=10,c=24$

$x_{vertex} = - \frac{10}{2} = - 5$ $x_(color(red)"vertex")=-10/2=-5$

$substitute this value into the equation for y$ $"substitute this value into the equation for y"$

$\Rightarrow y_{vertex} = {(- 5)}^{2} + 10 (- 5) + 24 = - 1$ $rArry_(color(red)"vertex")=(-5)^2+10(-5)+24=-1$

$\Rightarrow vertex = (- 5, - 1)$ $rArrcolor(magenta)"vertex "=(-5,-1)$

$axis of symmetry is x = - 5$ $"axis of symmetry is "x=-5$
graph{(y-x^2-10x-24)(y-1000x-5000)=0 [-10, 10, -5, 5]}

Answer 2 · 2017-08-25T10:40:41

The x intercepts are $x_{1} = - 6$ $x_1=-6$ and $x_{2} = - 4$ $x_2=-4$
The axis of symmetry is: $x = - 5$ $x=-5$
The vertex is $V = (- 5; - 1)$ $V=(-5;-1)$

Explanation:

In order to find the x intercepts, we will substitute $y = 0$ $y=0$ and then solve the equation $x^{2} + 10 x + 24 = 0$ $x^2+10x+24=0$ .
We will use the short quadratic formula $x = - \frac{b}{2} \pm \sqrt{{(\frac{b}{2})}^{2} - c}$ $x=-b/2+-sqrt((b/2)^2-c$ , useful when b is even and a=1 in the general form $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$

Then $x = - 5 \pm \sqrt{5^{2} - 24} = - 5 \pm 1$ $x=-5+-sqrt(5^2-24)=-5+-1$

and the x intercepts are $x_{1} = - 6$ $x_1=-6$ and $x_{2} = - 4$ $x_2=-4$

The axis of symmetry is:

$x = - \frac{b}{2 a} = - \frac{10}{2} = - 5$ $x=-b/(2a)=-10/2=-5$

We simply will get the second coordinate of the vertex by substituting x=-5 in the given F(x):

Then $V = (- 5; F (- 5)) = (- 5; 25 - 50 + 24) = (- 5; - 1)$ $V=(-5;F(-5))=(-5;25-50+24)=(-5;-1)$
graph{x^2+10x+24 [-10, 5, -5, 5]}

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Question #576da

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