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

1 Answer
Mar 28, 2018

Vertex of the parabola formed by color(red)(y=f(x)=5(x+2)^2+7 iscolor(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:

$y = f \left(x\right) = 5 {\left(x + 2\right)}^{2} + 7$

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

$\Rightarrow y = 5 \left[{x}^{2} + 4 x + 4\right] + 7$

$\Rightarrow y = 5 {x}^{2} + 20 x + 20 + 7$

$\therefore y = 5 {x}^{2} + 20 x + 27$

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 $- \frac{b}{2 a}$

$\Rightarrow - \frac{b}{2 a} = - \frac{20}{2 \cdot 5}$

$\Rightarrow - \frac{20}{10} = - 2$

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

$y = 5 {x}^{2} + 20 x + 27$

$y = 5 {\left(- 2\right)}^{2} + 20 \left(- 2\right) + 27$

$y = 20 - 40 + 27$

$y = 47 - 40$

$y = 7$

Hence, Vertex is at $\left(- 2 , 7\right)$

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 $y = f \left(x\right) = 5 {\left(x + 2\right)}^{2} + 7 = 0$

Subtract $7$ from both sides.

$\Rightarrow 5 {\left(x + 2\right)}^{2} + 7 - 7 = 0 - 7$

$\Rightarrow 5 {\left(x + 2\right)}^{2} + \cancel{7} - \cancel{7} = 0 - 7$

$\Rightarrow 5 {\left(x + 2\right)}^{2} = - 7$

Divide both sides by $5$

$\Rightarrow \frac{5 {\left(x + 2\right)}^{2}}{5} = - \frac{7}{5}$

$\Rightarrow \frac{\cancel{5} {\left(x + 2\right)}^{2}}{\cancel{5}} = - \frac{7}{5}$

$\Rightarrow {\left(x + 2\right)}^{2} = - \frac{7}{5}$

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 $x = 0$.

$\Rightarrow 5 {\left(0 + 2\right)}^{2} + 7$

$\Rightarrow {2}^{2} \cdot 5 + 7$

$\Rightarrow 27$

$\therefore y$-intercept is at $\left(0 , 27\right)$

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: