For which pair of functions is the vertex of k(x)7 units below the vertex of f(x)?

where a, b, and c are real numbers and \displaystyle a\ne 0a≠0. If \displaystyle a>0a>0, the parabola opens upward. If \displaystyle a<0a<0, the parabola opens downward. We can use the general form of a parabola to find the equation for the axis of symmetry.

The axis of symmetry is defined by \displaystyle x=-\frac{b}{2a}x=−​2a​​b​​. If we use the quadratic formula, \displaystyle x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}x=​2a​​−b±√​b​2​​−4ac​​​​​, to solve \displaystyle a{x}^{2}+bx+c=0ax​2​​+bx+c=0 for the x-intercepts, or zeros, we find the value of x halfway between them is always \displaystyle x=-\frac{b}{2a}x=−​2a​​b​​, the equation for the axis of symmetry.

Figure 4 shows the graph of the quadratic function written in general form as \displaystyle y={x}^{2}+4x+3y=x​2​​+4x+3. In this form, \displaystyle a=1,\text{ }b=4a=1, b=4, and \displaystyle c=3c=3. Because \displaystyle a>0a>0, the parabola opens upward. The axis of symmetry is \displaystyle x=-\frac{4}{2\left(1\right)}=-2x=−​2(1)​​4​​=−2. This also makes sense because we can see from the graph that the vertical line \displaystyle x=-2x=−2 divides the graph in half. The vertex always occurs along the axis of symmetry. For a parabola that opens upward, the vertex occurs at the lowest point on the graph, in this instance, \displaystyle \left(-2,-1\right)(−2,−1). The x-intercepts, those points where the parabola crosses the x-axis, occur at \displaystyle \left(-3,0\right)(−3,0) and \displaystyle \left(-1,0\right)(−1,0).

answer: 140

step-by-step explanation:

so just add all then it will be your answer but i am not pretty sur eif it is but hope this i am here to people