Question

What is the the vertex of `y = 2x^2 - 14x-5`?

Answer 1

`(x_("vertex") , y_("vertex")) -> (3 1/2, -29 1/2)`

Explanation:

Tony B

`color(blue)("Method 1")`

Given that the standard form for a quadratic equation is:

`ax^2+bx+c=0`

and: `color(white)(....)x =(-b+-sqrt(b^2-4ac))/(2a)`

Then you could use this to find the `x` intercepts and that `x_("vertex")` is half way between them. That is `color(blue)(-b/(2a))`
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`color(blue)("Method 2")`

`color(brown)("Use something that is similar to completing the square:")`
`color(green)("When you think about this, it is the same thing as method 1!")`

Write as: `y=2(x^2-14/2x)-5`

Now consider just the brackets

`color(blue)(x_("vertex")=) (-1/2)xx (-14/2)=+14/4 =color(blue)( +3 1/2 )`
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Having found `x_("vertex")` we can find the value of `y_("vertex")` by substituting for `x` in the original equation.

`y_("vertex")= 2x^2 -14x-5`

`y_("vertex")= 2(7/2)^2-14(7/2)-5`

`color(blue)(y_("vertex") =)49/2-49-5 = color(blue)(-29 1/2)`
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`(x_("vertex") , y_("vertex")) -> (3 1/2, -29 1/2)`