Question

I need help finding the x intercept and the vertex

Answer 1

So,

Given the function:

$y=-x^2+6x-5$

We're going to find the x-intercepts of the function as follows:

Step 1: Set the equation equal to zero.

$-x^2+6x-5=0$

Step 2: We could multiply both sides of the equation by -1 so we could factor after that:

$x^2-6x+5=0$

Now, we could factor this equation to solve for x and find the x-intercepts. To factor, we're going to find two numbers, whose sum gives us -6, and its product is 5. These numbers are -5 and -1, so we're going to group them in factors as follows:

$(x-5)(x-1)=0$

Solving for x, we notice that each factor could be zero, then:

$(x-5)(x-1)=0\to\begin{cases}x-5=0\to x=5 \\ x-1=0\to x=1\end{cases}$

Therefore, the x-intercepts of the function are x=1 and x=5.

To find the vertex, we could use the fact that every quadratic equation of the form:

$y=ax^2+bx+c$

Has a vertex in the following coordinates:

$V(x,y)=V(-\frac{b}{2a},-\frac{b^2}{4a}+c)$

What we're going to do is to identify the values of a,b, and c (Which are the coefficients of the terms of the equation) and then replace them in the previous equation so we could calculate the coordinates of the vertex. This is,

$y=-x^2+6x-5$

Where:

$\begin{gathered} a=-1 \\ b=6 \\ c=-5 \end{gathered}$

Replacing these values in the equation of the vertex, we got that:

$\begin{gathered} V(x,y)=V(\frac{-6}{2(-1)},\frac{-6^2}{4(-1)}-5) \\ \\ V(x,y)=V(\frac{-6}{-2},\frac{-36^{}}{-4}-5) \\ \\ V(x,y)=V(3,9-5) \\ V(x,y)=V(3,4) \end{gathered}$

Therefore, the coordinates of the vertex are (3,4).