Question

Sketch the graph of the function. $$y=5 x^{2}+4 x-5$$

Short answer

The graph of the function \(y=5x^2+4x-5\) is a parabola that opens upwards, with a vertex at \(-0.4, -5.8\) and x-intercepts at \(x=-1\) and \(x=1\).

Step-by-step solution

Determine the vertex

To find the vertex, use the formula \(h=-\frac{b}{2a}\) where \(a\) and \(b\) are coefficients from the equation \(y=ax^2+bx+c\). For the given function \(y=5x^2+4x-5\), \(a=5\) and \(b=4\). Plugging into the formula, vertex \(h=-\frac{b}{2a}=-\frac{4}{10}=-0.4\). Substitute the x-equivalent of vertex into the given equation to get the y-equivalent of the vertex. This step results in a vertex at \(-0.4, -5.8\)

Determine the x-intercepts

Set the equation equal to zero to find where the graph intersects the x-axis. These are the roots of the equation \(0=5x^2+4x-5\). To solve for \(x\), this may require factoring, completing the square, or the quadratic formula. Using the quadratic formula identified as \(x=\frac{-b \pm \sqrt{b^2-4ac} }{2a}\), the roots or x-intercepts are given by \(-1\) and \(1\)

Sketch the graph

Start by plotting the vertex point at \(-0.4, -5.8\). Then, plot the x-intercepts at \(x=-1\) and \(x=1\). Because the coefficient of \(x^2\) is positive, the graph opens upward. Connect these three points to form a general shape of the graph and sketch the final parabola. Ensure it passes through these critical points.