Chapter EP, Problem 5.7.5EP

To determine

To Find: Vertex form of given equation and identify vertex ,axis of symmetry,and direction of opening.

Answer to Problem 5.7.5EP

  $h=5,k=22$ ,Direction of opening ‘Downside’

Explanation of Solution

Given:

  $y=-x^2+10x-3$

Formula Used:

  $\begin{array}{l}\text{Vertex form of a quadratic function}\\ \Rightarrow f\left(x\right)=a{\left(x-h\right)}^2+k\\ \text{Where }\left(h,k\right)\text{ is the vertex of the parabola}\end{array}$

Calculation:

  $\begin{array}{l}y=-x^2+10x-3\\ \Rightarrow y-25=-\left(x^2-10x+25\right)-3\left(\text{making the function perfect square and substract both side by 25}\right)\\ \Rightarrow y=-{\left(x-5\right)}^2\text{+22 }\left(\text{Writing the function in the vertex form}\right)\\ \text{vertex are given below,}\\ \to h=5\\ \to k=22\end{array}$

  

Conclusion:

The vertex of the given function is $h=5,k=22$ and the direction of opening of parabola is Downside