Question

how many x-intercepts does the parabola with the following equation have?

( y = x^2 - 14x + 49 )

show your work here

hint: to add infinity (\\(\infty\\)), type \infinity\

enter your answer

Explanation:

Step1: Find x-intercepts by setting y=0

Set \( y = 0 \), so the equation becomes \( 0 = x^2 - 14x + 49 \).

Step2: Factor the quadratic equation

The quadratic \( x^2 - 14x + 49 \) can be factored as a perfect square trinomial. We know that \( (a - b)^2 = a^2 - 2ab + b^2 \). Here, \( a = x \) and \( 2ab = 14x \), so \( b = 7 \) (since \( 2\times x\times7 = 14x \)). Thus, \( x^2 - 14x + 49=(x - 7)^2 \). So the equation is \( 0=(x - 7)^2 \).

Step3: Solve for x

Taking the square root of both sides, we get \( x - 7 = 0 \), so \( x = 7 \). Since this is a repeated root (the square of a binomial gives a repeated solution), there is 1 distinct x - intercept (but with multiplicity 2). However, when counting the number of x - intercepts (distinct points where the graph crosses or touches the x - axis), we consider the number of distinct solutions. In this case, the parabola touches the x - axis at \( x = 7 \) (a single point), so there is 1 x - intercept (or we can say it has a repeated root, so the number of x - intercepts is 1, considering the geometric interpretation of the parabola touching the x - axis at one point).

Answer:

1