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All you need is in the photo and it's homework please don't do step by step

All you need is in the photo and it's homework please don't do step by step-example-1
User SeedyROM
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1 Answer

5 votes
5 votes

f(x)=x^2-4x-12

x-intercpets: value of x when f(x)=0. Use the quadratic formula:


\begin{gathered} x^2-4x-12=0 \\ \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \\ x=\frac{-(-4)\pm\sqrt[]{(-4)^2-4(1)(-12)}}{2(1)} \\ \\ x=\frac{4\pm\sqrt[]{64}}{2} \\ \\ x=(4\pm8)/(2) \\ \\ x_1=(4+8)/(2)=6 \\ \\ x_2=(4-8)/(2)=-2 \end{gathered}X-intercept: (6,0) and (-2,0)

y-intercept: value of f(x) when x=0:


\begin{gathered} f(0)=0^2-4(0)-12 \\ f(0)=-12 \end{gathered}Y-intercept: (0, -12)

Vertex: use the next foruma to find the x value in vertex:


\begin{gathered} x=-(b)/(2a) \\ \\ x=-((-4))/(2(1)) \\ \\ x=(4)/(2)=2 \end{gathered}

Use that value of x to find the coordinate in y for the vertex:


\begin{gathered} f(2)=2^2-4(2)-12 \\ f(2)=4-8-12 \\ f(2)=-16 \end{gathered}Vertex: (2, -16)

Axis of symmetry: value of x in the vertex.

Axis of simmetry: x=2

Maximim or minimum: if in the quadratic equation, the coefficient of the square x is possitive the parabola opens up and has a minimun value. If the coefficient of the square x is negative the parabola opens down and and has a maximum value.

Minimum

The minimum value is in the vertex (2 ,-16) is the value of the function f(x) in the vertex (the value of coordinate y):

Min value: -16Graph:

All you need is in the photo and it's homework please don't do step by step-example-1
User Einpoklum
by
2.3k points
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