Chapter 9.1, Problem 38PPS

To determine

To compute the vertex, the equation of the axis of symmetry and the y-intercept of the given function.

Answer to Problem 38PPS

Vertex is $\left(-2,-10\right)$ , equation of axis of symmetry is $x=-2$ and the y-intercept is $\left(0,10\right)$ .

Explanation of Solution

Given information :

The function provided is $y=5x^2+20x+10$ .

Formula used :

Formula to compute equation of the axis of symmetry $x=-\frac{b}{2\times a}$ . Here, $a=5$ and $b=20$ . ‘a’ and ‘b’ are coefficients $x^2$ and $x$ respectively.

To compute the vertex, use the value of x found in the axis of symmetry to compute the value of y.

For finding the y-intercept, substitute the value of x in the function with 0 and compute the value of y . y -intercept is always given by the notion (0,y) .

Calculation :

Axis of symmetry for the given function is

  $x=-\frac{b}{2\times a}$

Formula for axis of symmetry.

  $x=-\frac{20}{2\times \left(5\right)}$

Putting the values of ‘a’ and ‘b’ .

  $=-\frac{20}{10}$

Simplifying

  $x=-2$ is the equation of axis of symmetry.

Vertex can be found out by putting the value of x computed in the axis of symmetry in the original function. This will give a value of y . These two coordinates of x and y would be the point where the vertex is.

  $=5{\left(-2\right)}^2+20\left(-2\right)+10$

Putting the value of $x=-3$ in the original function.

  $=20-40+10$

Simplifying the expression.

  $=-10$

Thus the vertex is $\left(-2,-10\right)$ .

y-intercept is computed by substituting the value of x in the equation by 0.

  $=5{\left(0\right)}^2+20\left(0\right)+10$

Simplifying

  $=10$

Hence the point of y-intercept is $\left(0,10\right)$ .