Determine the vertex of y = 3x^2- 6x – 5.

Consider the vertex form of a parabola. Substitute the values of a a and b b into the formula d=b2a d = b 2 a . Simplify the right side. Cancel the common factor of 6 6 and 2 2 .

Step-by-step explanation:

the x value of the vertex in ax^2+bx+c is -b/2athe x value of the vertex in ax^2+bx+c is -b/2ay value is just sub for xthe x value of the vertex in ax^2+bx+c is -b/2ay value is just sub for x3x^2-6x+5the x value of the vertex in ax^2+bx+c is -b/2ay value is just sub for x3x^2-6x+5x value of vertex is -(-6)/(2*3)=6/6=1the x value of the vertex in ax^2+bx+c is -b/2ay value is just sub for x3x^2-6x+5x value of vertex is -(-6)/(2*3)=6/6=1sub backthe x value of the vertex in ax^2+bx+c is -b/2ay value is just sub for x3x^2-6x+5x value of vertex is -(-6)/(2*3)=6/6=1sub backy=3(1)^2-6(1)+5the x value of the vertex in ax^2+bx+c is -b/2ay value is just sub for x3x^2-6x+5x value of vertex is -(-6)/(2*3)=6/6=1sub backy=3(1)^2-6(1)+5y=3(1)-6+5the x value of the vertex in ax^2+bx+c is -b/2ay value is just sub for x3x^2-6x+5x value of vertex is -(-6)/(2*3)=6/6=1sub backy=3(1)^2-6(1)+5y=3(1)-6+5y=3-1the x value of the vertex in ax^2+bx+c is -b/2ay value is just sub for x3x^2-6x+5x value of vertex is -(-6)/(2*3)=6/6=1sub backy=3(1)^2-6(1)+5y=3(1)-6+5y=3-1y=2the x value of the vertex in ax^2+bx+c is -b/2ay value is just sub for x3x^2-6x+5x value of vertex is -(-6)/(2*3)=6/6=1sub backy=3(1)^2-6(1)+5y=3(1)-6+5y=3-1y=2vertex is (1,2)

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