Question

How do you find the vertex and intercepts for `y = 5(x+2)^2 + 7`?

Answer 1

Vertex of the parabola formed by `color(red)(y=f(x)=5(x+2)^2+7` is`color(blue)((-2,7)` .

x-intercept: Does not exist.

**y-intercept: ** `color(blue)((0,27)`

Explanation:

Standard form of a quadratic function is `color(blue)(y =ax^2+bx+c`.

The parabola will open up , if the the coefficient of `x^2` term is positive.

The parabola will open down , if the the coefficient of `x^2` term is negative.

Let us consider the quadratic function given to us:

`y=f(x)= 5(x+2)^2+7`

Using the algebraic identity `color(red)((a+b)^2 -= a^2 + 2ab + b^2`,

`rArr y=5[x^2+4x+4]+7`

`rArr y = 5x^2+20x+20+7`

`:. y = 5x^2+20x+27`

We have seen that

Standard form of a quadratic function is `color(blue)(y =ax^2+bx+c`.

Note that, `color(red)(a=5; b=20 and c=27`

To find the x-coordinate of the Vertex, use the formula `-b/(2a)`

`rArr -b/(2a) = -20/(2*5)`

`rArr -20/10 = -2`

To find the y-coordinate of the vertex, substitute `x = -2` in

`y = 5x^2+20x+27`

`y= 5(-2)^2+20(-2)+27`

`y = 20-40+27`

`y=47-40`

`y=7`

Hence, **Vertex ** is at `(-2, 7)`

Since, the coefficient of the `x^2` term is positive, the parabola opens up.

x-intercept is a point on the graph where `y=0`.

Solve `y=f(x)=5(x+2)^2+7=0`

Subtract `7` from both sides.

`rArr 5(x+2)^2+7-7=0-7`

`rArr 5(x+2)^2+cancel 7- cancel 7=0-7`

`rArr 5(x+2)^2=-7`

Divide both sides by `5`

`rArr (5(x+2)^2)/5=-7/5`

`rArr (cancel 5(x+2)^2)/cancel 5=-7/5`

`rArr (x+2)^2 = -7/5`

Observe that `color(blue)[[g(x) ]^2` cannot be negative for `color(red)(x in RR`.

y-intercept is the point on the graph where `x=0`.

`rArr 5(0+2)^2+7`

`rArr 2^2*5+7`

`rArr 27`

`:. y`-intercept is at `(0,27)`

Hence,

Vertex of the parabola formed by:

`color(red)(y=f(x)=5(x+2)^2+7`

is `color(blue)((-2,7)` .

x-intercept: Does not exist.

**y-intercept: ** `color(blue)((0,27)`

An image of the graph is available below: