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Write the quadratic function in standard form and sketch its graph. Identify the vertex, axis of symmetry, and \(x\) -intercept(s). $$f(x)=x^{2}+34 x+289$$

Short Answer

Expert verified
The standard form of the quadratic function is \(f(x) = (x + 17)^2\). The vertex of the parabola is (-17, 0), the axis of symmetry is \(x = -17\), and the x-intercept is (-17, 0).

Step by step solution

01

Convert to standard form

We'll start by rewriting \(f(x) = x^2 + 34x + 289\) in standard (vertex) form, which is \(f(x) = a(x-h)^2 + k\), where \((h, k)\) is the vertex of the parabola. The given quadratic function is a perfect square trinomial, that is \(f(x) = (x + 17)^2\). This is now in standard form.
02

Identify the vertex and axis of symmetry

We will identify the vertex of the parabola by comparing our equation to the vertex form. The vertex is given by the point \((h, k)\), thus our vertex is \((-17, 0)\). The axis of symmetry is a vertical line that passes through the vertex. Therefore, our axis of symmetry is \(x = -17\).
03

Find the x-intercepts

The x-intercepts are the points where the parabola intersects the x-axis. This happens when the y-value (or \(f(x)\) ) is 0. Hence, to find the x-intercepts, we set \(f(x) = 0\). Hence, \(0 = (x + 17)^2\). Solving for x gives \(x = -17\). So, the x-intercept is \((-17, 0)\). However, please consider that this quadratic function has exactly one x-intercept, as it's a perfect square trinomial, and it touches but does not cross the x-axis.

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