Question

How do you find the vertex and intercepts for `y=x^2+10x+21`?

Answer 1

vertex `-> (x,y)->(-5,-4)`

`y_("intercept")=21`

`x_("intercept")=-3 " and " -7`

Explanation:

Given: `color(blue)(y=x^2+10x+21)`

`color(brown)("To find the vertex")`

Consider the +10 from +10x

`color(brown)(x_("vertex")) = (-1/2) xx (+10)color(brown)(=-5)`

Substitute the found value of `x_("vertex")` in the original equation to find `y_("vertex")`

`y_("vertex")=(-5)^2+10(-5)+21`

`color(brown)(y_("vertex")=-4`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Each intercept is when the plotted line crosses one of the axis. That is: when `x=0` the plot crosses `y_("intercept")` and when `y=0` the plot crosses `x_("intercept")`

`color(brown)("So to find "y_("intercept") " write the equation as:")`

`y_("intercept")=0^2+10(0)+21`

`color(brown)(y_("intercept")=+21)`

You will observe from this that it is the value of the constant.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(brown)("So to find "x_("intercept") " write the equation as:")`

`0=x^2+10x+21`

I notice that `3xx7=21` and `3+7=10`

so we can factorise giving:

`color(blue)(0=(x+3)(x+7))color(green)( -> x^2+7x+3x+21) = x^2+10x+21`

So we have to make each of the bracketed parts 0 to give `y=0`

So for `(x+3)=0` the value of `x=-3`
and for `(x+7)=0` the value of `x=-7`

So `color(brown)(x_("intercept") = -7 " and " -3)`