How do you find the vertex and axis of symmetry, and then graph the parabola given by: $f (x) = 2 x^{2} - 6 x - 5$ $f(x)=2x^2-6x-5$ ?

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Answer 1 · 2015-09-26T01:43:08

The x-coordinate of the vertex is equal to $- \frac{b}{2 a}$ $-b/(2a)$

The general format of a quadratic equation is:

$f (x) = a x^{2} + b x + c$ $f(x) = ax^2+bx+c$

So, for this problem:

$a = 2$ $a = 2$
$b = - 6$ $b=-6$
$c = - 5$ $c=-5$

The vertex (x,y) has an x value:
$x = - \frac{b}{2 a} = - \frac{- 6}{2 \cdot 2} = 1.5$ $x = -b/(2a) = -(-6)/(2*2) = 1.5$

To find the y value of the vertex, simply plug the x value (1.5) back into the equation:

$y = 2 ({1.5}^{2}) - 6 (1.5) - 5 = - 9.5$ $y = 2(1.5^2)-6(1.5)-5 = -9.5$

So the vertex is $(1.5, - 9.5)$ $(1.5, -9.5)$

The axis of symmetry travels vertical through the x value of the vetex:

axis of symmetry: $x = 1.5$ $x = 1.5$

In order to graph the function, simply set up an x-y table and plug some x values {-5,-4,-3,-2,-1,0,1,2,3,4,5} into the equation to solve for y:

graph{2x^2-6x-5 [-18.3, 21.7, -10.8, 9.2]}

Hope that helped

How do you find the vertex and axis of symmetry, and then graph the parabola given by: f(x)=2x2−6x−5f(x)=2x^2-6x-5?