The vertex and [tex]x-[/tex]intercepts of the graph of [tex]y= x^2 - 6x - 7[/tex] will be [tex](3,-16)[/tex] and [tex](7,-1)[/tex], respectively.
What are vertex and [tex]x-[/tex]intercepts of the graph?
A vertex in a quadratic equation is the minimum or maximum point of the equation and an [tex]x-[/tex]intercept of a graph is the point at which the graph intersects the [tex]x-[/tex]axis.
Vertex form, [tex]y=a(x-h)^2+k[/tex]
We have,
[tex]y= x^2 - 6x - 7[/tex]
Now,
Find the [tex]x-[/tex]intercept and to find [tex]x-[/tex]intercept put [tex]y=0[/tex] ;
i.e. [tex]y= x^2 - 6x - 7[/tex]
[tex]0= x^2 - 6x - 7[/tex]
[tex]x^2 - 6x - 7=0[/tex]
Now using Mid-term split method find value of [tex]x[/tex];
[tex]x^2 - 7x+1x - 7=0[/tex]
[tex]x(x - 7)+1(x - 7)=0[/tex]
[tex](x - 7)(x +1)=0[/tex]
[tex](x - 7)=0[/tex]⇒[tex]x=7[/tex]
[tex](x +1)=0[/tex]⇒[tex]x=-1[/tex]
So, these are the [tex]x-[/tex]intercept [tex](7,-1)[/tex] of the expression.
Now find the vertex using the above given formula;
[tex]y=a(x-h)^2+k[/tex]
Rewrite in vertex form
[tex]y= x^2 - 6x - 7[/tex]
Add and subtract [tex](-3)^2[/tex] in given expression;
[tex]y= x^2 - 6x + (-3)^2 -7 - (-3)^2[/tex]
[tex]y = x^2 - 6x + 9 -7 - 9[/tex]
[tex]y = x^2 - 6x + 9 -16[/tex]
Now convert quadratic part of equation in [tex](a-b)^2[/tex];
[tex]y = (x-3)^2-16[/tex]
Now compare it with [tex]y=a(x-h)^2+k[/tex]
We get, [tex]h=3,\ k=-16[/tex]
So, these are the vertex [tex](3,-16)[/tex] of expression.
Hence, we can say that the vertex and [tex]x-[/tex]intercepts of the graph of [tex]y= x^2 - 6x - 7[/tex] will be [tex](3,-16)[/tex] and [tex](7,-1)[/tex], respectively.
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